2.4: Factoring Trinomials
- Page ID
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- Factor trinomials of the form \(x^2+bx+c\)
- Factor trinomials of the form \(ax^2+bx+c\) using trial and error
- Factor trinomials of the form \(ax^2+bx+c\) using the ‘\(ac\)’ method
- Factor perfect square trinomials
- Factor differences of squares
You may choose to learn
- Factor using substitution
- Factor sums and differences of cubes
Factor Trinomials of the Form \(x^2+bx+c\)
You have already learned how to multiply binomials using FOIL. Now you’ll need to “undo” this multiplication. To factor the trinomial means to start with the product, and end with the factors.
To figure out how we would factor a trinomial of the form \(x^2+bx+c\), such as \(x^2+5x+6\) and factor it to \((x+2)(x+3)\), let’s start with two general binomials of the form \((x+m)\) and \((x+n)\).
| \((x+m)(x+n)\) | |
| Foil to find the product. | \(x^{2}+m x+n x+m n\) |
| Factor the GCF from the middle terms. | \(x^{2}+(m+n) x+m n\) |
| Our trinomial is of the form \(x^2+bx+c\). | \(\overbrace{x^{2}+(m+n) x+m n}^{\color{red}x^{2}+b x+c}\) |
This tells us that to factor a trinomial of the form \(x^2+bx+c\), we need two factors \((x+m)\) and \((x+n)\) where the two numbers \(m\) and \(n\) multiply to \(c\) and add to \(b\).
Factor: \(x^2+11x+24\).
- Answer
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Let’s summarize the steps we used to find the factors.
- Write the factors as two binomials with first terms x. \(\quad \begin{array} {l} x^2+bx+c \\ (x\quad)(x\quad) \end{array} \)
- Find two numbers \(m\) and \(n\) that
- multiply to \(c\), \(m·n=c\)
- add to \(b\), \(m+n=b\)
- Use \(m\) and \(n\) as the last terms of the factors. \(\quad (x+m)(x+n)\)
- Check by multiplying the factors.
In the first example, all terms in the trinomial were positive. What happens when there are negative terms? Well, it depends which term is negative. Let’s look first at trinomials with only the middle term negative.
How do you get a positive product and a negative sum? We use two negative numbers.
Factor: \(y^2−11y+28\).
- Answer
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Again, with the positive last term, \(28\), and the negative middle term, \(−11y\), we need two negative factors. Find two numbers that multiply \(28\) and add to \(−11\).
\(\begin{array} {ll} &y^2−11y+28 \\ \text{Write the factors as two binomials with first terms }y. &( y \quad )( y \quad ) \\ \text{Find two numbers that: multiply to }28\text{ and add to }−11.\end{array}\)Factors of \(28\) Sum of factors \(−1,\space −28\)
\(−2,\space −14\)
\(−4,\space −7\)\(−1+(−28)=−29\)
\(−2+(−14)=−16\)
\(−4+(−7)=−11^∗\)\(\begin{array} {ll} \text{Use }−4,\space −7\text{ as the last terms of the binomials.} &(y−4)(y−7) \\ \text{Check:} & \\ \hspace{30mm} (y−4)(y−7) & \\ \hspace{25mm} y^2−7y−4y+28 & \\ \hspace{30mm} y^2−11y+28\checkmark & \end{array} \)
Now, what if the last term in the trinomial is negative? Think about FOIL. The last term is the product of the last terms in the two binomials. A negative product results from multiplying two numbers with opposite signs. You have to be very careful to choose factors to make sure you get the correct sign for the middle term, too.
How do you get a negative product and a positive sum? We use one positive and one negative number.
When we factor trinomials, we must have the terms written in descending order—in order from highest degree to lowest degree.
Factor: \(2x+x^2−48\).
- Answer
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\(\begin{array} {ll} &2x+x^2−48 \\ \text{First we put the terms in decreasing degree order.} &x^2+2x−48 \\ \text{Factors will be two binomials with first terms }x. &(x\quad)(x\quad) \end{array} \)
Factors of −48 Sum of factors \(−1,\space 48\)
\(−2,\space 24\)
\(−3,\space 16\)
\(−4,\space 12\)
\(−6,\space 8\)\(−1+48=47\)
\(−2+24=22\)
\(−3+16=13\)
\(−4+12=8\)
\(−6+8=2^∗\)\(\begin{array} {ll} \text{Use }−6,\space 8\text{ as the last terms of the binomials.} &(x−6)(x+8) \\ \text{Check:} & \\ \hspace{30mm} (x−6)(x+8) & \\ \hspace{25mm} x^2−6q+8q−48 & \\ \hspace{30mm} x^2+2x−48\checkmark & \end{array} \)
Factor: \(y^2+14y+49\).
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Again, with the positive last term, \(49\), and the popsitive middle term, \(14y\), we need two positive factors. Find two numbers that multiply \(49\) and add to \(14\).
\(\begin{array} {ll} &y^2+14y+49 \\ \text{Write the factors as two binomials with first terms }y. &( y \quad )( y \quad ) \\ \text{Find two numbers that: multiply to }49\text{ and add to }14.\end{array}\)Factors of \(49\) Sum of factors \(1,\space 49\)
\(7,\space 7\)\(1+49)=50\)
\(7+7=14\)
\(\begin{array} {ll} \text{Use }7,\space 7\text{ as the last terms of the binomials.} &(y+7)(y+7) \\ \text{Check:} & \\ \hspace{30mm} (y+7)(y+7) & \\ \hspace{25mm} y^2+7y+7y+49 & \\ \hspace{30mm} y^2+14y+49\checkmark & \end{array} \)
Factor: \(y^2−12y+36\).
- Answer
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Again, with the positive last term, \(36\), and the negative middle term, \(−12y\), we need two negative factors. Find two numbers that multiply \(36\) and add to \(−12\).
\(\begin{array} {ll} &y^2−12y+36 \\ \text{Write the factors as two binomials with first terms }y. &( y \quad )( y \quad ) \\ \text{Find two numbers that: multiply to }36\text{ and add to }−12.\end{array}\)Factors of \(28\) Sum of factors \(−1,\space −36\)
\(−2,\space −18\)
\(−3,\space −12\)\(−4,\space −9\)
\(−6,\space −6\)
\(−1+(−36)=−37\)
\(−2+(−18)=−20\)
\(−3+(−12)=−15\)\(-4+(-9)=-13\)
\(-6+(-6)=-12\)
\(\begin{array} {ll} \text{Use }−6,\space −6\text{ as the last terms of the binomials.} &(y−6)(y−6) \\ \text{Check:} & \\ \hspace{30mm} (y−6)(y−6) & \\ \hspace{25mm} y^2−6y−6y+36 & \\ \hspace{30mm} y^2−12y+36\checkmark & \end{array} \)
Some trinomials are prime. The only way to be certain a trinomial is prime is to list all the possibilities and show that none of them work.
Factor: \(u^2−9uv−12v^2\).
- Answer
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We need \(u\) in the first term of each binomial and \(v\) in the second term. The last term of the trinomial is negative, so the factors must have opposite signs.
\(\begin{array} {ll} &u^2−9uv−12v^2 \\ \text{Note that the first terms are }u,\text{ last terms contain }v. &(u\quad v)(u\quad v) \\ \text{Find the numbers that multiply to }−12\text{ and add to }−9. & \end{array} \)Factors of \(−12\) Sum of factors \(1,−12\)
\(−1,12\)
\(2,−6\)
\(−2,6\)
\(3,−4\)
\(−3,4\)\(1+(−12)=−11\)
\(−1+12=11\)
\(2+(−6)=−4\)
\(−2+6=4\)
\(3+(−4)=−1\)
\(−3+4=1\)Note there are no factor pairs that give us \(−9\) as a sum. The trinomial is prime.
Factor Trinomials of the Form \(ax^2+bx+c\) using the “\(ac\)” Method
Another way to factor trinomials of the form \(ax^2+bx+c\) is the “\(ac\)” method. (The “\(ac\)” method is sometimes called the grouping method.) The “\(ac\)” method is actually an extension of the methods you used in the last section to factor trinomials with leading coefficient one. This method is very structured (that is step-by-step), and it always works!
Factor using the “\(ac\)” method: \(6x^2+7x+2\).
- Answer
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The “\(ac\)” method is summarized here.
- Factor any GCF.
- Find the product \(ac\).
- Find two numbers \(m\) and \(n\) that:
\(\begin{array} {ll} \text{Multiply to }ac &m·n=a·c \\ \text{Add to }b &m+n=b \\ &ax^2+bx+c \end{array} \) - Split the middle term using \(m\) and \(n\). \(ax^2+mx+nx+c\)
- Factor by grouping.
- Check by multiplying the factors.
Factor: \(64x^2-1\).
- Answer
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Is there a greatest common factor? No The trinomial has a leading coefficient that is not \(1\). \(64x^2-1\) Find the product \(ac\). \(ac=-64\) Find two numbers that multiply to \(ac\) \((-8)(8)=64\) and add to \(b\). \(−8+8=0\) \(64x^2+8x-8x-1\). \(8x(8x+1)-1(8x+1)\) Factor the trinomial by grouping. \((8x-1)(8x+1)\) Check by multiplying all three factors.
\(\hspace{50mm} (8x-1)(8x+1)\)\(\hspace{45mm} 64x^2+8x−8x-1)\)
\(\hspace{49mm} 64x^2-1\checkmark\)
Factor using the ‘“\(ac\)” method: \(10y^2−55y+70\).
- Answer
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Is there a greatest common factor? Yes. The GCF is \(5\). 
Factor it. 
The trinomial inside the parentheses has a
leading coefficient that is not \(1\).
Find the product \(ac\). \(ac=28\) Find two numbers that multiply to \(ac\) \((−4)(−7)=28\) and add to \(b\). \(−4(−7)=−11\) Split the middle term. 
Factor the trinomial by grouping. 
Check by multiplying all three factors.
\(\hspace{50mm} 5(y−2)(2y−7)\)\(\hspace{45mm} 5(2y^2−7y−4y+14)\)
\(\hspace{48mm} 5(2y^2−11y+14)\)
\(\hspace{49mm} 10y^2−55y+70\checkmark\)
We have seen that some binomials and trinomials result from special products—squaring binomials and multiplying conjugates. If you learn to recognize these kinds of polynomials, you can use the special products patterns to factor them much more quickly.
Factor Perfect Square Trinomials
We have already seem some examples of perfect squares, but if you desire more practice, you may work through these sections. Some trinomials are perfect squares. They result from multiplying a binomial times itself. We squared a binomial using the Binomial Squares pattern in a previous chapter.
The trinomial \(9x^2+24x+16\) is called a perfect square trinomial. It is the square of the binomial \(3x+4\).
In this chapter, you will start with a perfect square trinomial and factor it into its prime factors. You could factor this trinomial using the methods described in the last section, since it is of the form \(ax^2+bx+c\). But if you recognize that the first and last terms are squares and the trinomial fits the perfect square trinomials pattern, you will save yourself a lot of work. Here is the pattern—the reverse of the binomial squares pattern.
If \(a\) and \(b\) are real numbers
\[a^2+2ab+b^2=(a+b)^2\nonumber\]
\[a^2−2ab+b^2=(a−b)^2\nonumber\]
To make use of this pattern, you have to recognize that a given trinomial fits it. Check first to see if the leading coefficient is a perfect square, \(a^2\). Next check that the last term is a perfect square, \(b^2\). Then check the middle term—is it the product, \(2ab\)? If everything checks, you can easily write the factors.
Factor: \(9x^2+12x+4\).
- Answer
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The sign of the middle term determines which pattern we will use. When the middle term is negative, we use the pattern \(a^2−2ab+b^2\), which factors to \((a−b)^2\).
The steps are summarized here.
\(\begin{array} {lllll} \textbf{Step 1.} &\text{Does the trinomial fit the pattern?} &\quad &\hspace{7mm} a^2+2ab+b^2 &\hspace{7mm} a^2−2ab+b^2 \\ &\text{Are the first and last terms perfect squares?} &\quad & &\\ &\text{Write them as squares.} &\quad &\hspace{5mm}(a)^2\hspace{16mm} (b)^2 &\hspace{6mm}(a)^2\hspace{16mm} (b)^2 \\ &\text{Check the middle term. Is it }2ab? &\quad &\hspace{12mm} {\,}^{\searrow}{\,}_{2·a·b}{\,}^{\swarrow} &\hspace{12mm} {\,}^{\searrow}{\,}_{2·a·b}{\,}^{\swarrow} \\ \textbf{Step 2.} &\text{Write the square of the binomial.} &\quad &\hspace{13mm} (a+b)^2 &\hspace{13mm} (a−b)^2 \\ \textbf{Step 3.} &\text{Check by multiplying.} & & & \end{array}\)
We’ll work one now where the middle term is negative.
Factor: \(81y^2−72y+16\).
- Answer
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The first and last terms are squares. See if the middle term fits the pattern of a perfect square trinomial. The middle term is negative, so the binomial square would be \((a−b)^2\).
\(81 y^{2}-72 y+16\) Are the first and last terms perfect squares? 
Check the middle term. 
Does it match \((a−b)^2\)? Yes. 
Write as the square of a binomial. \((9 y-4)^{2}\) Check by multiplying:
\[(9y−4)^2\nonumber\]\[(9y)^2−2·9y·4+4^2\nonumber\]\[81y^2−72y+16\checkmark\nonumber\]
Remember the first step in factoring is to look for a greatest common factor. Perfect square trinomials may have a GCF in all three terms and it should be factored out first. And, sometimes, once the GCF has been factored, you will recognize a perfect square trinomial.
Factor: \(100x^2y−80xy+16y\).
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\(100 x^{2} y-80 x y+16 y\) Is there a GCF? Yes, \(4y\), so factor it out. \(4 y\left(25 x^{2}-20 x+4\right)\) Is this a perfect square trinomial? Verify the pattern. 
Factor. \(4 y(5 x-2)^{2}\) Remember: Keep the factor 4y in the final product. Check:
\[4y(5x−2)^2\nonumber\]\[4y[(5x)2−2·5x·2+22]\nonumber\]\[4y(25x2−20x+4)\nonumber\]100x2y−80xy+16y\checkmark\]
Factor Differences of Squares
The other special product you saw in the previous chapter was the Product of Conjugates pattern. You used this to multiply two binomials that were conjugates. Here’s an example:
A difference of squares factors to a product of conjugates.
If \(a\) and \(b\) are real numbers,
Remember, “difference” refers to subtraction. So, to use this pattern you must make sure you have a binomial in which two squares are being subtracted.
\(\begin{array} {llll} \textbf{Step 1.} &\text{Does the binomial fit the pattern?} &\qquad &\hspace{5mm} a^2−b^2 \\ &\text{Is this a difference?} &\qquad &\hspace{2mm} \text{____−____} \\ &\text{Are the first and last terms perfect squares?} & & \\ \textbf{Step 2.} &\text{Write them as squares.} &\qquad &\hspace{3mm} (a)^2−(b)^2 \\ \textbf{Step 3.} &\text{Write the product of conjugates.} &\qquad &(a−b)(a+b) \\ \textbf{Step 4.} &\text{Check by multiplying.} & & \end{array}\)
It is important to remember that sums of squares do not factor into a product of binomials. There are no binomial factors that multiply together to get a sum of squares. After removing any GCF, the expression \(a^2+b^2\) is prime!
The next example shows variables in both terms.
Factor: \(144x^2−49y^2\).
- Answer
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\(\begin{array} {lll} &\quad &144x^2−49y^2 \\ \text{Is this a difference of squares? Yes.} &\quad &(12x)^2−(7y)^2 \\ \text{Factor as the product of conjugates.} &\quad &(12x−7y)(12x+7y) \\ \text{Check by multiplying.} &\quad &(12x−7y)(12x+7y) \\ \text{Check by multiplying.} &\quad & \\ &\quad & \\ &\quad & \\ \hspace{14mm} (12x−7y)(12x+7y) &\quad & \\ \hspace{21mm} 144x^2−49y^2\checkmark &\quad & \end{array}\)
As always, you should look for a common factor first whenever you have an expression to factor. Sometimes a common factor may “disguise” the difference of squares and you won’t recognize the perfect squares until you factor the GCF.
For the Mathematically Curious: Factor Using Substitution
Sometimes a trinomial does not appear to be in the \(ax^2+bx+c\) form. However, we can often make a thoughtful substitution that will allow us to make it fit the \(ax^2+bx+c\) form. This is called factoring by substitution. It is standard to use \(u\) for the substitution.
In the \(ax^2+bx+c\), the middle term has a variable, \(x\), and its square, \(x^2\), is the variable part of the first term. Look for this relationship as you try to find a substitution.
Factor by substitution: \(x^4−4x^2−5\).
- Answer
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The variable part of the middle term is \(x^2\) and its square, \(x^4\), is the variable part of the first term. (We know \((x^2)^2=x^4)\). If we let \(u=x^2\), we can put our trinomial in the \(ax^2+bx+c\) form we need to factor it.
\(x^4−4x^2−5\) Rewrite the trinomial to prepare for the substitution. \((x^2)^2−4(x^2)-5\) Let \(u=x^2\) and substitute. \((u)^2−4(u)-5\) Factor the trinomial. \((u+1)(u-5)\) Replace \(u\) with \(x^2\). \((x^2+1)(x^2-5)\) Check:
\(\begin{array} {l} \hspace{37mm} (x^2+1)(x^2−5) \\ \hspace{35mm}x^4−5x^2+x^2−5 \\ \hspace{40mm}x^4−4x^2−5\checkmark\end{array}\)
Sometimes the expression to be substituted is not a monomial.
Factor by substitution: \((x−2)^2+7(x−2)+12\)
- Answer
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The binomial in the middle term, \((x−2)\) is squared in the first term. If we let \(u=x−2\) and substitute, our trinomial will be in \(ax^2+bx+c\) form.
Rewrite the trinomial to prepare for the substitution. 
Let \(u=x−2\) and substitute. 
Factor the trinomial. 
Replace \(u\) with \(x−2\). 
Simplify inside the parentheses. 
This could also be factored by first multiplying out the \((x−2)^2\) and the \(7(x−2)\) and then combining like terms and then factoring. Most students prefer the substitution method
For the Mathematically Curious: Factor Sums and Differences of Cubes
There is another special pattern for factoring, one that we did not use when we multiplied polynomials. This is the pattern for the sum and difference of cubes. We will write these formulas first and then check them by multiplication.
\[a^3+b^3=(a+b)(a^2−ab+b^2\nonumber\]
\[a^3−b^3=(a−b)(a^2+ab+b^2)\nonumber\]
We’ll check the first pattern and leave the second to you.
| \(\color{red}(a+b) \color{black} \left(a^{2}-a b+b^{2}\right)\) | |
| Distribute. | \(\color{red}a \color{black}\left(a^{2}-a b+b^{2}\right)+ \color{red}b \color{black}\left(a^{2}-a b+b^{2}\right)\) |
| Multiply. | \(a^{3}-a^{2} b+a b^{2}+a^{2} b-a b^{2}+b^{3}\) |
| Combine like terms. | \(a^{3}+b^{3}\) |
\[a^3+b^3=(a+b)(a^2−ab+b^2\nonumber\]\[a^3−b^3=(a−b)(a^2+ab+b^2)\nonumber\]
The two patterns look very similar, don’t they? But notice the signs in the factors. The sign of the binomial factor matches the sign in the original binomial. And the sign of the middle term of the trinomial factor is the opposite of the sign in the original binomial. If you recognize the pattern of the signs, it may help you memorize the patterns.
The trinomial factor in the sum and difference of cubes pattern cannot be factored.
It be very helpful if you learn to recognize the cubes of the integers from 1 to 10, just like you have learned to recognize squares. We have listed the cubes of the integers from 1 to 10 in Table.
| n | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |
|---|---|---|---|---|---|---|---|---|---|---|
| \(n^3\) | 1 | 8 | 27 | 64 | 125 | 216 | 343 | 512 | 729 | 1000 |
Factor: \(x^3+64\).
- Answer
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- Does the binomial fit the sum or difference of cubes pattern?
Is it a sum or difference?
Are the first and last terms perfect cubes? - Write them as cubes.
- Use either the sum or difference of cubes pattern.
- Simplify inside the parentheses.
- Check by multiplying the factors.
Factor: \(27u^3−125v^3\).
- Answer
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\(27 u^{3}-125 v^{3}\) This binomial is a difference. The first and last
terms are perfect cubes.Write the terms as cubes. 
Use the difference of cubes pattern. 
Simplify. 
Check by multiplying. We’ll leave the check to you.
In the next example, we first factor out the GCF. Then we can recognize the sum of cubes.
Factor: \(6x^3y+48y^4\).
- Answer
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\(6 x^{3} y+48 y^{4}\) Factor the common factor. \(6 y\left(x^{3}+8 y^{3}\right)\) This binomial is a sum The first and last
terms are perfect cubes.Write the terms as cubes. 
Use the sum of cubes pattern. 
Simplify. 
Check:
To check, you may find it easier to multiply the sum of cubes factors first, then multiply that product by 6y.6y. We’ll leave the multiplication for you.
The first term in the next example is a binomial cubed.
Factor: \((x+5)^3−64x^3\).
- Answer
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\((x+5)^{3}-64 x^{3}\) This binomial is a difference. The first and
last terms are perfect cubes.Write the terms as cubes. 
Use the difference of cubes pattern. 
Simplify. \((x+5-4 x)\left(x^{2}+10 x+25+4 x^{2}+20 x+16 x^{2}\right)\) \((-3 x+5)\left(21 x^{2}+30 x+25\right)\) Check by multiplying. We’ll leave the check to you.
For the mathematically curious: Factor Trinomials of the form ax2 + bx + c using Trial and Error
Our next step is to factor trinomials whose leading coefficient is not 1, trinomials of the form \(ax^2+bx+c\).
Remember to always check for a GCF first! Sometimes, after you factor the GCF, the leading coefficient of the trinomial becomes \(1\) and you can factor it by the methods we’ve used so far. Let’s do an example to see how this works.
Factor completely: \(4x^3+16x^2−20x\).
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\(\begin{array} {lll} \text{Is there a greatest common factor?} &\qquad &4x^3+16x^2−20x \\ \quad \text{Yes, }GCF=4x.\text{ Factor it.} & &4x(x^2+4x−5) \\ & & \\ & & \\ \text{Binomial, trinomial, or more than three terms?} & & \\ \quad \text{It is a trinomial. So “undo FOIL.”} & &4x(x\quad)(x\quad) \\ & & \\ & & \\ \text{Use a table like the one shown to find two numbers that} & &4x(x−1)(x+5) \\ \text{multiply to }−5\text{ and add to }4. & & \\ & & \\ & & \end{array} \)
Factors of \(−5\) Sum of factors \(−1,5\)
\(1,−5\)\(−1+5=4^∗\)
\(1+(−5)=−4\)\(\begin{array} {l} \text{Check:}\\ \hspace{27mm}4x(x−1)(x+5) \\ \hspace{25mm} 4x(x^2+5x−x−5) \\ \hspace{30mm} 4x(x^2+4x−5) \\ \hspace{25mm} 4x^3+16x2−20x\checkmark \end{array} \)
What happens when the leading coefficient is not \(1\) and there is no GCF? There are several methods that can be used to factor these trinomials. First we will use the Trial and Error method.
Let’s factor the trinomial \(3x^2+5x+2\).
From our earlier work, we expect this will factor into two binomials.
\[3x^2+5x+2\nonumber\]\[(\quad)(\quad)\nonumber\]
We know the first terms of the binomial factors will multiply to give us \(3x^2\). The only factors of \(3x^2\) are \(1x,\space 3x\). We can place them in the binomials.
Check: Does \(1x·3x=3x^2\)?
We know the last terms of the binomials will multiply to \(2\). Since this trinomial has all positive terms, we only need to consider positive factors. The only factors of \(2\) are \(1\) and \(2\). But we now have two cases to consider as it will make a difference if we write \(1\), \(2\) or \(2\), \(1\).
Which factors are correct? To decide that, we multiply the inner and outer terms.
Since the middle term of the trinomial is \(5x\), the factors in the first case will work. Let’s use FOIL to check.
\[(x+1)(3x+2)\nonumber\]\[3x^2+2x+3x+2\nonumber\]\[3x^2+5x+2\checkmark\nonumber\]
Our result of the factoring is:
\[3x^2+5x+2\nonumber\]\[(x+1)(3x+2)\nonumber\]
Factor completely using trial and error: \(3y^2+22y+7\).
- Answer
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Remember, when the middle term is negative and the last term is positive, the signs in the binomials must both be negative.
Factor completely using trial and error: \(6b^2−13b+5\).
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The trinomial is already in descending order. 
Find the factors of the first term. 
Find the factors of the last term. Consider the signs.
Since the last term, \(5\), is positive its factors must both be
positive or both be negative. The coefficient of the
middle term is negative, so we use the negative factors.
Consider all the combinations of factors.
\(6b^2−13b+5\) Possible factors Product \((b−1)(6b−5)\) \(6b^2−11b+5\) \((b−5)(6b−1)\) \(6b^2−31b+5\) \((2b−1)(3b−5)\) \(6b^2−13b+5^∗\) \((2b−5)(3b−1)\) \(6b^2−17b+5\) \(\begin{array} {ll} \text{The correct factors are those whose product} & \\ \text{is the original trinomial.} &(2b−1)(3b−5) \\ \text{Check by multiplying:} & \\ \hspace{50mm} (2b−1)(3b−5) & \\ \hspace{47mm} 6b^2−10b−3b+5 & \\ \hspace{50mm} 6b^2−13b+5\checkmark & \end{array} \)
When we factor an expression, we always look for a greatest common factor first. If the expression does not have a greatest common factor, there cannot be one in its factors either. This may help us eliminate some of the possible factor combinations.
Remember to look for a GCF first and remember if the leading coefficient is negative, so is the GCF.
Factor completely using trial and error: \(−10y^4−55y^3−60y^2\).
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Notice the greatest common factor, so factor it first. 
Factor the trinomial. 
Consider all the combinations.
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\(\begin{array} {ll} \text{The correct factors are those whose product} & \\ \text{is the original trinomial. Remember to include} & \\ \text{the factor }−5^y2. &−5y^2(y+4)(2y+3) \\ \text{Check by multiplying:} & \\ \hspace{50mm} −5y^2(y+4)(2y+3) & \\ \hspace{45mm} −5y^2(2y^2+8y+3y+12) & \\ \hspace{47mm}−10y^4−55y^3−60y^2\checkmark & \end{array} \)
Key Concepts
- How to factor trinomials of the form \(x^2+bx+c\).
- Write the factors as two binomials with first terms x. \(\quad \begin{array} (l) x^2+bx+c \\ (x\quad)(x\quad)\end{array}\)
- Find two numbers \(m\) and \(n\) that
\(\begin{array} {ll} \text{multiply to} &c,\space m·n=c \\ \text{add to} &b,\space m+n=b\end{array}\) - Use \(m\) and \(n\) as the last terms of the factors. \(\qquad (x+m)(x+n)\)
- Check by multiplying the factors.
- Strategy for Factoring Trinomials of the Form \(x^2+bx+c\): When we factor a trinomial, we look at the signs of its terms first to determine the signs of the binomial factors.
For trinomials of the form: \(x^2+bx+c = (x+m)(x+n)\)
When \(c\) is positive, \(m\) and \(n\) must have the same sign (and this will be the sign of \(b\)).
Examples: \(x^2+5x+6=(x+2)(x+3)\), \(x^2−6x+8 = (x−4)(x−2)\)
When \(c\) is negative, \(m\) and \(n\) have opposite signs. The larger of \(m\) and \(n\) will have the sign of \(b\).
Examples: \(x^2+x−12=(x+4)(x−3)\), \(x^2−2x−15=(x−5)(x+3)\)
Notice that, in the case when \(m\) and \(n\) have opposite signs, the sign of the one with the larger absolute value matches the sign of \(b\).
- How to factor trinomials of the form \(ax^2+bx+c\) using trial and error.
- Write the trinomial in descending order of degrees as needed.
- Factor any GCF.
- Find all the factor pairs of the first term.
- Find all the factor pairs of the third term.
- Test all the possible combinations of the factors until the correct product is found.
- Check by multiplying.
- How to factor trinomials of the form \(ax^2+bx+c\) using the “\(ac\)” method.
- Factor any GCF.
- Find the product \(ac\).
- Find two numbers \(m\) and \(n\) that:
\(\begin{array} {ll} \text{Multiply to }ac. &m·n=a·c \\ \text{Add to }b. &m+n=b \\ &ax^2+bx+c\end{array}\) - Split the middle term using \(m\) and \(n\). \(\quad ax^2+mx+nx+c\)
- Factor by grouping.
- Check by multiplying the factors.
- Perfect Square Trinomials Pattern: If a and b are real numbers,
\[\begin{array} {l} a^2+2ab+b^2=(a+b)^2 \\ a^2−2ab+b^2=(a−b)^2\end{array} \nonumber\]
- How to factor perfect square trinomials.
\(\begin{array} {lllll} \textbf{Step 1.} &\text{Does the trinomial fit the pattern?} &\quad &\hspace{7mm} a^2+2ab+b^2 &\hspace{7mm} a^2−2ab+b^2 \\ &\text{Are the first and last terms perfect squares?} &\quad & &\\ &\text{Write them as squares.} &\quad &\hspace{5mm}(a)^2\hspace{16mm} (b)^2 &\hspace{6mm}(a)^2\hspace{16mm} (b)^2 \\ &\text{Check the middle term. Is it }2ab? &\quad &\hspace{12mm} {\,}^{\searrow}{\,}_{2·a·b}{\,}^{\swarrow} &\hspace{12mm} {\,}^{\searrow}{\,}_{2·a·b}{\,}^{\swarrow} \\ \textbf{Step 2.} &\text{Write the square of the binomial.} &\quad &\hspace{13mm} (a+b)^2 &\hspace{13mm} (a−b)^2 \\ \textbf{Step 3.} &\text{Check by multiplying.} & & & \end{array}\)
- Difference of Squares Pattern: If a,ba,b are real numbers,
- How to factor differences of squares.
\(\begin{array} {llll} \textbf{Step 1.} &\text{Does the binomial fit the pattern?} &\qquad &\hspace{5mm} a^2−b^2 \\ &\text{Is this a difference?} &\qquad &\hspace{2mm} \text{____−____} \\ &\text{Are the first and last terms perfect squares?} & & \\ \textbf{Step 2.} &\text{Write them as squares.} &\qquad &\hspace{3mm} (a)^2−(b)^2 \\ \textbf{Step 3.} &\text{Write the product of conjugates.} &\qquad &(a−b)(a+b) \\ \textbf{Step 4.} &\text{Check by multiplying.} & & \end{array}\) - Sum and Difference of Cubes Pattern
\(\begin{array} {l} a^3+b3=(a+b)(a^2−ab+b^2) \\ a^3−b^3=(a−b)(a^2+ab+b^2) \end{array} \) - How to factor the sum or difference of cubes.
- Does the binomial fit the sum or difference of cubes pattern?
Is it a sum or difference?
Are the first and last terms perfect cubes? - Write them as cubes.
- Use either the sum or difference of cubes pattern.
- Simplify inside the parentheses
- Check by multiplying the factors.
- Does the binomial fit the sum or difference of cubes pattern?