Page 1.6: Factoring Polynomials
Suppose we were asked to simplify \( 7x^2yz(2y^2 + 3xz^2) \). How would we do this? Quite simply, we would distribute to obtain \( 14x^2 y^3 z + 21x^3 yz^3 \).
The question we will now consider is essentially the reverse. That is, starting with \( 14x^2 y^3 z + 21x^3 yz^3 \) how can we go back to \( 7x^2yz(2y^2 + 3xz^2) \)?
The answer is that we simply check the greatest common factor/divisor of each corresponding part. Notice that:
1) \( gcd(14, 21) = 7 \)
2) \( gcd( x^2, x^3) = x^2 \)
3) \( gcd( y^2, y) = y \)
4) \( gcd( z, z^3) = z \)
So each term in \( 14x^2 y^3 z + 21x^3 yz^3 \) shares a factor of \( 7x^2yz \). Hence
\[ \frac{14x^2 y^3 z + 21x^3 yz^3 }{ 7x^2yz} = \frac{14x^2 y^3 z }{7x^2yz} + \frac{21x^3 yz^3}{7x^2yz} = 2y^2 + 3xz^2 \nonumber\ \]
That is, \[ \frac{14x^2 y^3 z + 21x^3 yz^3 }{ 7x^2yz} = 2y^2 + 3xz^2 \nonumber\ \]
Cross multiplying yields \( 14x^2 y^3 z + 21x^3 yz^3 = 7x^2yz(2y^2 + 3xz^2) \). With enough practice, we will be able to do this in one line.
Factor \( 8x^2y^3 + 16xy^4 - 32x^5y \).
- Answer
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\(8xy( xy^2 + 2y^3 - 4x^4 ) \)
Factor \( 48ab^2 - 64 a^3b^5\).
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\( 16ab^2(3-4a^2b^3) \)
When factoring, we should keep an eye out for special cases. For example, one such case we should look out for is when we are asked to factor the difference of two squares.
More concretely, consider the expression \( 100x^2 - 25 \) which is a difference of two squares. Observe that \( 100x^2 - 25 = (100x + 5)(100x - 5) \).
More generally, \(a^2 - b^2) = (a-b)(a+b) \) where \(a\) and \(b\) may represent real numbers or algebraic expressions. A list of some special cases are provided below.
Theorem: Let \(a\) and \(b\) denote real numbers or algebraic expressions. Then
1) \(a^2 - b^2) = (a+b)(a-b) \)
2) \(a^2 + 2ab + b^2 = (a+b)^2 \)
3) \(a^2 - 2ab + b^2 = (a-b)^2 \)
Factor the following:
1) \( 49x^2 - 81 \)
2) \( 4x^2-36 \)
Answer
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1) \( 49x^2 - 81 = (7x+9)(7x-9) \)
2) \( 4x^2-36 = (2x+6)(2x-6) = 2(x+3)2(x-3) = 4(x+3)(x-3) \)
We will often have to factor a polynomial, especially quadratics. Recall from the last section that a quadratic is a polynomial of the form \( a_2 x^2 + a_1 x + a_0 \). However, we will commonly denote a quadratic as \(ax^2 + bx + c \). Allow us to now discuss some techniques on how to factor a quadratic.
To factor \(ax^2 + bx + c \), we perform the following steps:
1) First multiply \(a\) by \(c\).
2) Search for two factors of \(ac\) whose sum is \(b\).
3) Rewrite \(b\) as a sum in the quadratic and then factor by grouping.
Factor \( x^2 - 5x + 6 \)
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1) We first multiply the \(a\) term with the \(c\) term. Doing so yields 6.
2) We now must find factors of \(6\) which sums to \(-5\). We see that \(-2\) and \(-3\) does the trick.
3) \begin{align*} x^2 - 5x + 6 &= x^2 -2x - 3x + 6 \\ &= (x^2 - 2x) + (-3x + 6) \\ &= x(x-2) - 3(x-2) \\ &= (x-2)(x-3) \end{align*}
In your previous studies, you may have seen a more "simple" way to factor quadratics when \(a = 1\).
To factor \(x^2 + bx + c \) the AC Method can be simplified to the following:
1) Search for two factors of \(c\) whose sum is \(b\). Let us call these factors \(r_1\) and \(r_2\).
2) Then \(x^2 + bx + c = (x + r_1) (x + r_2) \).
Factor \( x^2 - 5x + 6 \)
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1) Since \(a = 1\) our first step should be finding factors of \(6\) which sums to \(-5\). We see that \(-2\) and \(-3\) does the trick. So \(r_1 = -2 \) and \( r_2 = -3 \).
2) Then \( x^2 - 5x + 6 = (x -2) (x-3) \)
Again, this latter technique only works when \(a= 1\).
Factor \( 2x^2 - 5x -3 \)
- Answer
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1) We first multiply the \(a\) term with the \(c\) term. Doing so yields \(-6 \).
2) We now must find factors of \(-6\) which sums to \(-5\). We see that \(-6\) and \(1\) does the trick.
3) \begin{align*} 2x^2 - 5x -3 &= 2x^2 - 6x + x -3 \\ &= (2x^2 - 6x) + (x -3) \\ &= 2x (x-3) + 1(x-3) \\ &= (x-3)(2x+1) \end{align*}
Factor:
1) \(3x^2 + x - 4 \)
2) \(4x^2 + 5x - 6 \)
3) \(2x^2 - 7x + 6 \)
4) \(5x^2 - 9x -2 \)
5) \( x^2 +2x - 24 \)
6) \(x^2 + 8x + 15 \)
- Answer
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1) \(3x^2 + x - 4 = (3x+4)(x-1) \)
2) \(4x^2 + 5x - 6 = (x+2)(4x-3) \)
3)\(2x^2 - 7x + 6 = (2x-3)(x-2) \)
4) \(5x^2 - 9x -2 = (5x+1)(x-2) \)
5) \( x^2 +2x - 24 = (x+6)(x-4) \)
6) \(x^2 + 8x + 15 (x+3)(x+5) \)
Solving Quadratic Equations
We will now turn our attention to solving quadratic equations. We first note the following theorem.
Theorem: Suppose \(a\) and \(b\) are real numbers or algebraic expressions. If \(ab=0\) then either \(a = 0\) or \(b = 0 \).
With The Zero Product Property, we can begin to solve equations such as quadratics. When we say solve an equation we mean we can find all values for the variable which will make the equation true.
To solve \(ax^2 + bx + c = 0 \) we first attempt to factor the quadratic. Then we set each factor equal to zero and solve.
Solve the equation \( 2x^2 - 5x -3 = 0 \)
- Answer
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1) We first multiply the \(a\) term with the \(c\) term. Doing so yields \(-6 \).
2) We now must find factors of \(-6\) which sums to \(-5\). We see that \(-6\) and \(1\) does the trick.
3) \begin{align*} 2x^2 - 5x -3 &= 2x^2 - 6x + x -3 \\ &= (2x^2 - 6x) + (x -3) \\ &= 2x (x-3) + 1(x-3) \\ &= (x-3)(2x+1) \end{align*}
4) \begin{align*} 2x^2 - 5x -3 &= 0 \\ (x-3)(2x+1) &= 0 \\ x-3 = 0 ~ &\text{or} ~ 2x+1=0 \\ x=3 ~ &\text{or} x = - \frac{1}{2} \end{align*}
Solve the equation \( x^2 - 7x = 0 \)
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1) We first factor this as \(x(x-7) = 0 \).
2) We now set each factor to 0 and solve. Hence either \(x=0\) or \(x=7\).
Solve the equation \( x^2 + 2x - 24 = 0 \)
- Answer
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1) We first factor \(x^2 + 2x - 24 \) as \( (x+6)(x-4) \).
2) We now set each factor to 0 and solve. Hence either \(x=-6\) or \(x=4\).
Let us now modify the above problem ever so slightly. Suppose we were asked to solve the equation \( x^2 + 2x - 25 = 0 \). Take a few moments and try our AC technique!
The above request shows that not every quadratic equation can be solved via the AC Method. Allow us to play around with the above equation and let us see what we can do. For now, do not worry as to why we are performing the following steps. Instead, just follow along:
\begin{align*} x^2 + 2x - 25 &= 0 && \text{adding 25 to both sides yields} \\ x^2 + 2x &= 25 && \text{adding 1 to both sides yields} \\ x^2 + 2x + 1 &= 26 && \text{factoring the left hand side yields} \\ (x+1)^2 &= 26 && \text{taking the square roots of both sides yields} \\ \sqrt{(x+1)^2} &= \sqrt{26} && \text{recall that} \sqrt{a^2} = |a| \\ |x+1| &= \sqrt{26} && \text{so either}~ x+1 = \sqrt{26} ~ \text{or} ~ x+1 = - \sqrt{26} \end{align*}
If \( x+1 = \sqrt{26} \) then \(x = 1 - \sqrt{26} \) and if \( x+1 = - \sqrt{26} \) then \(x = -1 - \sqrt{26} \).
Allow us to now reflect on the above process. We initially had an equation of the form \(ax^2 + bx +c = 0 \) which we could not factor. We then subtracted \(c\) from both sides to obtain \(ax^2 + bx = -c \). Then, we added in a "magical" number, \(m\) to both sides and obtained \(ax^2 + bx +m = -c + m\). The reason this number was "magical" is because the left hand side then turned into a perfect square!
The process outlined above is called completing the square . Completing the square is a method which can be used to solve all quadratics - factorable and non-factorable!
One question you may have is how do we determine this magical number \(m\)? Allow us to state the process of completing the square which answers this question.
1) Given an equation of the form \( ax^2 + bx + c = 0 \) subtract \(c\) from both sides.
2) Divide both sides by \(a\).
3) Add \(m\) to both sides where \(m = \bigg(dfrac{\text{coefficient on x}}{2}\bigg)^2 \).
4) The left hand side is now a perfect square and so we may solve the equation.
Solve \(x^2 - 4x + 2 = 0\).
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1) Subtracting 2 from both sides yields \(x^2 - 4x = -2\).
2) Dividing both sides by 1 does not affect the equation.
3) \( m = \bigg( \frac{-4}{2} \bigg)^2 = 4 \) and so we add 4 to both sides of the equation.
4) Adding 4 to both sides of the equation yields \(x^2 - 4x +4 = 2\). The left hand side is precisely \( (x-2)^2 \) and so we have to solve \( (x-2)^2 = 2\). Taking the square root of both sides yields \( |x-2| = \sqrt{2} \) and so either \( x-2 = sqrt{2} \) or \( x-2 = - \sqrt{2} \). Hence \(x = 2 \pm \sqrt{2} \).
Solve \(5x^2 - 9x - 2 = 0\).
- Answer
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1) Adding 2 to both sides yields \(5x^2 - 9x = 2 \).
2) Dividing both sides by 5 yields \( x^2 - \frac{9}{5} x = \frac{2}{5} \).
3) \( m = \bigg( \frac{ \frac{9}{5} }{2} \bigg)^2 = \frac{81}{100} \) and so we add \( \frac{81}{100} \) to both sides of the equation.
4) Adding \( \frac{81}{100} \) to both sides of the equation yields \( x^2 - \frac{9}{5} x + \frac{81}{100} = \frac{2}{5} + \frac{81}{100} \). The left hand side is precisely \( (x- \frac{9}{10})^2 \) and so we have to solve \( (x- \frac{9}{10})^2 = \frac{121}{100} \). Taking the square root of both sides yields \( |x- \frac{9}{10}| = \sqrt{ \frac{121}{100}} \) and so either \( x- \frac{9}{10} = \sqrt{ \frac{121}{100}} \) or \( x- \frac{9}{10} = - \sqrt{ \frac{121}{100}}\). Hence \(x = \frac{9}{10} \pm \frac{121}{100}\) which yields \(x = 2 \) or \( x = - \frac{1}{5} \).
Solve \(2x^2 - 10x - 3 = 0\).
- Answer
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\( x = \frac{5}{2} \pm \sqrt{\frac{31}{4}} \).
Solve \(ax^2 + bx + c = 0\). You may assume that \(a \neq 0 \).
- Answer
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1) Subtracting \(c\) from both sides yields \(ax^2 + bx = -c \).
2) Dividing both sides by \(a\) yields \( x^2 + \frac{b}{a}x = - \frac{c}{a} \).
3) \( m = \bigg( \frac{ \frac{b}{a} }{2} \bigg)^2 = \frac{b^2}{4a^2} \) and so we add \( \frac{b^2}{4a^2} \) to both sides of the equation.
4) Adding \( \frac{b^2}{4a^2} \) to both sides of the equation yields \( x^2 + \frac{b}{a} x + \frac{b^2}{4a^2} = \frac{b^2}{4a^2} - \frac{c}{a} \). Simplifying the right hand side via a common denominator yields \( x^2 + \frac{b}{a} x + \frac{b^2}{4a^2} = \frac{b^2}{4a^2} - \frac{4ac}{4a^2} \) or equivalently \( x^2 + \frac{b}{a} x + \frac{b^2}{4a^2} = \frac{b^2 - 4ac}{4a^2} \).
Now we will simplify the left hand side and solve. Doing so yields \( (x+ \frac{b}{2a} )^2 = \frac{b^2 - 4ac}{4a^2} \).
Taking the square root of both sides yields \[\sqrt{(x+ \frac{b}{2a} )^2} = \sqrt{\frac{b^2 - 4ac}{4a^2}} \nonumber\ \]. Recalling that \( \sqrt{a^2} = |a| \) the above line becomes
\[ |x+ \frac{b}{2a}| = \sqrt{\frac{b^2 - 4ac}{4a^2}} \nonumber\ \]
Hence \[ x+ \frac{b}{2a}= \pm \sqrt{\frac{b^2 - 4ac}{4a^2}} \nonumber\ \] and so \[ x= - \frac{b}{2a} \pm \sqrt{\frac{b^2 - 4ac}{4a^2}} \nonumber\ \] Simplifying the right hand side yields \[ x= - \frac{b}{2a} \pm \frac{\sqrt{b^2-4ac}}{2a} \nonumber\ \] and adding the two fractions yields \[ x = \frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \]
which is precisely the quadratic formula.
The Quadratic Formula
Theorem: The solution(s) to \(ax^2 + bx + c = 0 \) where \(a \neq 0 \) are given by \[ x = \frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \nonumber \]
Using the quadratic formula, solve \(5x^2 - 9x - 2 = 0\).
- Answer
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\[ x = \frac{-b\pm\sqrt[]{b^2-4ac}}{2a} = \frac{9\pm\sqrt[]{(-9)^2-4(5)(-2)}}{2(5)} = \frac{9\pm\sqrt{121}}{10} = frac{9\pm 11}{10} \nonumber \] Hence \(x = 2 \) or \( x = - \frac{1}{5} \).
Using the quadratic formula, solve \(3x^2 - 4x - 2 = 0\).
- Answer
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\[ x = \frac{2 \pm \sqrt{10}}{3} \nonumber\ \].
Note that in the quadratic formula, we refer to \( b^2 - 4ac \) as the discriminant . Whenever the discriminant is negative, the quadratic will have no real solutions. For instance, one may check that the solutions to \( 3x^2 - 4x + 2 = 0 \) do not exist among the real numbers.