6.7: Factoring Trinomials with Leading Coefficient 1
- Page ID
- 49378
Method
Let’s consider the product of the two binomials \((x+4)\) and \((x+7)\).
Notice that the first term in the resulting trinomial comes from the product of the first terms in the binomials: \(x \cdot x=x^2\). The last term in the trinomial comes from the product of the last terms in the binomials: \(4 \cdot 7=28\). The middle term comes from the addition of the outer and inner products: \(7x+4x=11x\). Also, notice that the coefficient of the middle term is exactly the sum of the last terms in the binomials: \(4+7=11\).
The problem we’re interested in is that given a trinomial, how can we find the factors? When the leading coefficient (the coefficient of the quadratic term) is 1, the observations we made above lead us to the following method of factoring.
- Write two sets of parentheses: ( ) ( ).
- Place a binomial into each set of parentheses. The first term of each binomial is a factor of the first term of the trinomial.
- Determine the second terms of the binomials by determining the factors of the third term that when added together yield the coefficient of the middle term.
Sample Set A
Factor the following trinomials.
\(x^2 + 5x + 6\)
1. Write two sets of parentheses: ( ) ( ).
2. Place the factors of \(x^2\) into the first position of each set of parentheses:
(\(x\) ) ( \(x\) )
3. The third term of the trinomial is \(6\). We seek two numbers whose
(a) product is \(6\)
(b) sum is \(5\)
The required numbers are \(3\) and \(2\). Place \(+3\) and \(+2\) into the parentheses.
\(x^2 + 5x + 6 = (x+3)(x+2)\)
The factorization is complete. We'll check to be sure:
\(\begin{array}{flushleft}
(x+3)(x+2)&=&x^2 + 2x + 3x + 6\\
&=&x^2 + 5x + 6
\end{array}\)
\(y^2 - 2y - 24\)
1. Write two sets of parentheses: ( ) ( ).
2. Place the factors of \(y^2\) into the first position of each set of parentheses:
(\(y\) ) ( \(y\) )
3. The third term of the trinomial is \(-24\). We seek two numbers whose
(a) product is \(-24\) and
(b) sum is \(-2\)
The required numbers are \(-6\) and \(4\). Place \(-6\) and \(+4\) into the parentheses.
\(y^2 - 2y - 24 = (y-6)(y+4)\)
The factorization is complete. We'll check to be sure.
\(\begin{array}{flushleft}
(y-6)(y+4)&=&y^2+4y-6y-24\\
&=&y^2-2y-24
\end{array}\)
Notice that the other combinations of \(-24\) (some of which are \(-2, 12; 3, -8; -4,6\)), do not work. For example,
\(\begin{array}{flushleft}
(y-2)(y+12)&=&y^2+10y-24\\
(y+3)(y-8)&=&y^2-5y-24\\
(y-4)(y+6)&=&y^2+2y-24
\end{array}\)
In all of these equations, the middle terms are incorrect.
\(a^2 - 11a + 30\)
1. Write two sets of parentheses: ( ) ( ).
2. Place the factors of \(a^2\) into the first position of each set of parentheses:
(\(a\) ) ( \(a\) )
3. The third term of the trinomial is \(+30\). We seek two numbers whose:
(a) product is \(30\) and
(b) sum is \(-11\).
The required numbers are \(-5\) and \(-6\). Place \(-5\) and \(-6\) into the parentheses.
\(a^2 - 11a + 30 = (a-5)(a-6)\)
The factorization is complete. We'll check to be sure
\(\begin{array}{flushleft}
(a-5)(a-6)&=&a^2 - 6a - 5a + 30\\
&=&a^2 - 11a + 30
\end{array}\)
\(3x^2 - 15x - 42\)
Before we begin, let's recall the most basic rule of factoring: factor out common monomial factors first. Notice that \(3\) is the greatest common monomial factor of every term. Factor out \(3\).
\(3x^2 - 15x - 42 = 3(x^2 - 5x - 14)\)
Now we can continue.
1. Write two sets of parentheses: \(3\)( ) ( ).
2. Place the factors of \(x^2\) into the first position of each set of parentheses:
\(3\)(\(x\) ) ( \(x\) )
3. The third term of the trinomial is \(-14\). We seek two numbers whose
(a) product is \(-14\) and
(b) sum is \(-5\).
The required numbers are \(-7\) and \(2\). Place \(-7\) and \(+2\) into the parentheses.
\(3x^2 - 15x - 42 = 3(x-7)(x+2)\)
The factorization is complete. We'll check to be sure.
\(\begin{array}{flushleft}
3(x-7)(x+2)&=&3(x^2+2x-7x-14)\\
&=&3(x^2-5x-14)\\
&=&3x^2 - 15x - 42
\end{array}\)
Practice Set A
Factor, if possible, the following trinomials.
\(k^2+8k+15\)
- Answer
-
\((k+3) (k+5)\)
\(y^2+7y−30\)
- Answer
-
\((y+10) (y−3)\)
\(m^2+10m+24\)
- Answer
-
\((m+6) (m+4)\)
\(m^2−10m+16\)
- Answer
-
\((m−8) (m−2)\)
Factoring Hints
Factoring trinomials may take some practice, but with time and experience, you will be able to factor much more quickly.
There are some clues that are helpful in determining the factors of the third term that when added yield the coefficient of the middle term.
Look at the sign of the last term:
- If the sign is positive, we know that the two factors must have the same sign, since \((+) (+)=(+)\) and \((−) (−) = (+)\). The two factors will have the same sign as the sign of the middle term.
- If the sign is negative, we know that two factors must have opposite signs, since \((+) (−)=(−)\) and \((−) (+) = (−)\).
Sample Set B
Factor \(x^2 - 7x + 12\)
1. Write two sets of parentheses: ( ) ( ).
2. The third term of the trinomial is \(+12\). The sign is potitive, so the two factors of \(12\) we are looking for must have the same sign. They will have the sign of the middle term. The sign of the middle term is negative, so both factors of \(12\) are negative. THey are \(-12\) and \(-1\), \(-6\) and \(-2\), or \(-4\) and \(-3\). Only the factors \(-4\) and \(-3\) add to \(-7\), so \(-4\) and \(-3\) are the proper factors of \(12\) to be used.
\(x^2 - 7x + 12 = (x-4)(x-3)\)
Practice Set B
Factor, if possible, the following trinomials.
\(4k^2+32k+28\)
- Answer
-
\(4(k+7) (k+1)\)
\(3y^4+24y^3+36y^2\)
- Answer
-
\(3y^2(y+2) (y+6)\)
\(x^2−xy−6y^2\)
- Answer
-
\((x+2y) (x−3y)\)
\(−5a^5b−10a^4b^2+15a^3b^3\)
- Answer
-
\(−5a^3b(a+3b) (a−b)\)
Exercises
For the following problems, factor the trinomials when possible.
\(x^2+4x+3\)
- Answer
-
\((x+3)(x+1)\)
\(x^2+6x+8\)
\(x^2+7x+12\)
- Answer
-
\((x+3)(x+4)\)
\(x^2+6x+5\)
\(y^2+8y+12\)
- Answer
-
\((y+6)(y+2)\)
\(y^2−5y+6\)
\(y^2−5y+4\)
- Answer
-
\((y−4)(y−1)\)
\(a^2+a−6\)
\(a^2+3a−4\)
- Answer
-
\((a+4)(a−1)\)
\(x^2+4x−21\)
\(x^2−4x−21\)
- Answer
-
\((x−7)(x+3)\)
\(x^2+7x+12\)
\(y^2+10y+16\)
- Answer
-
\((y+8)(y+2)\)
\(x^2+6x−16\)
\(y^2−8y+7\)
- Answer
-
\((y−7)(y−1)\)
\(y^2−5y−24\)
\(a^2+a−30\)
- Answer
-
\((a+6)(a−5)\)
\(a^2−3a+2\)
\(a^2−12a+20\)
- Answer
-
\((a−10)(a−2)\)
\(y^2−4y−32\)
\(x^2+13x+42\)
- Answer
-
\((x+6)(x+7)\)
\(x^2+2x−35\)
\(x^2+13x+40\)
- Answer
-
\((x+5)(x+8)\)
\(y^2+6y−27\)
\(b^2+15b+56\)
- Answer
-
\((b+8)(b+7)\)
\(3a^2+24a+36\)
(Hint: Always search for a common factor.)
\(4x^2+12x+8\)
- Answer
-
\(4(x+2)(x+1)\)
\(2a^2−18a+40\)
\(5y^2−70y+440\)
- Answer
-
\(5(y^2−14y+88)\)
\(6x^2−54x+48\)
\(x^3+6x^2+8x\)
- Answer
-
\(x(x+4)(x+2)\)
\(x^3−8x^2+15x\)
\(x^4+9x^3+14x^2\)
- Answer
-
\(x^2(x+7)(x+2)\)
\(2a^3+12a^2+10a\)
\(4a^3−40a^2+84a\)
- Answer
-
\(4a(a−7)(a−3)\)
\(3xm^2+33xm+54x\)
\(2y^2n^2−10y^2n−48y^2\)
- Answer
-
\(2y^2(n−8)(n+3)\)
\(4x^4−42x^3+144x^2\)
\(y^5+13y^4+42y^3\)
- Answer
-
\(y^3(y+6)(y+7)\)
\(4x^2a^6−48x^2a^5+252x^2a^4\)
Exercises for Review
Factor \(6xy+2ax−3ay−a^2\).
- Answer
-
\((2x−a)(3y+a)\)
Factor \(8a^2−50\).
Factor \(4x^2+17x−15.\)
- Answer
-
\((4x−3)(x+5)\)