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8.3: Factoring and Finding Polynomial Solutions (Zeroes)

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    There are several ways to find solutions of polynomials which are trinomials of the form \(ax^2 + bx + c = 0\). These solutions are also called the real zeroes of polynomials.

    1. Trial and Check Factoring Method: With this method, the goal is to create two binomials that when multiplied together, result in the given trinomial. This method can be very difficult when the given trinomial has large values of \(a\) and \(c\). Once the factoring is complete find all real zeroes by using the zero factor property and setting each factor equal to \(0\) and solve for \(x\).
    2. Factor by Grouping Factoring Method: With this method, the goal is to create four terms by splitting the middle term into two terms, whose coefficients have a product of \(a ∗ c\) and have a sum of \(b\). The order of the center terms does not matter. Once the four terms are created, pair the first two terms with parentheses, pair the second two terms with parentheses, and factor out the GCF from both pairs. The resultant repeated binomial is one factor, and the GCF factors combine to make the second binomial. This is the easiest method to use on any factorable trinomial of the form \(ax^2 + bx + c\), but may have a bit of a learning curve. Once the factoring is complete find all real zeroes by using the zero factor property and setting each factor equal to \(0\) and solve for \(x\).
    3. The Quadratic Formula: The Quadratic Formula can be used to find the real zeroes of a factorable trinomial. Please see the Table of Contents to find the section explaining how to use the Quadratic Formula.
    Example 8.3.1

    Factor the expressions using any of the methods discussed in this section (these example problems will demonstrate the Factor by Grouping method):

    1. \(4x^2 − 3x − 10\)
    2. \(8x^2 − 2x − 3\)
    3. \(12x − 14x^3 + 22x^2\)
    4. \(\dfrac{(x^2 + 1)^2 (−2) + (2x)2(x^2 + 1)(2x)}{(x^2 + 1)^4}\)
    5. \(\dfrac{(2x + 1)^{\frac{1}{2}} − (x + 2)(2x + 1)^{-\frac{1}{2}}}{2x + 1}\)
    Solution
    1. \(\begin{array} &&4x^2 − 3x − 10 &\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\text{Example problem} \\ &4x^2 − 3x − 10 &\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\text{Product \(ac\) is \(4∗(−10) = −40\), Sum is \(b = −3\). To use factor by grouping, } \\ & &\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\text{two middle terms are needed that multiply to a product of \(−40\) and add to a sum of \(−3\).} \\ & &\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; \text{\(8\) and \(5\) are good candidates; Since the product must be negative, one of these values must be negative.} \\ & &\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; \text{\(−8\) and positive \(5\) will work, because their product is \(−40\) and their sum is \(−3\).} \\ &4x^2 − 8x + 5x − 10 &\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\text{Four terms, the sum of the two middle terms is the original middle term, \(−3x\)} \\ &(4x^2 − 8x) + (5x − 10) &\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\text{Create pairs of terms} \\ &4x(x − 2) + 5(x − 2) &\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\text{Factor out the GCF from each pair- a repeated binomial factor is always present} \\ &(4x + 5)(x − 2) &\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\text{Solution. Be sure to check by FOIL.} \end{array}\)
    1. \(\begin{array} &&8x^2 − 2x − 3 &\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\text{Example problem} \\ &8x^2 − 2x − 3 &\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\text{Product \(ac\) is \(8∗(−3) = −24\), Sum is \(b = −2\). To use factor by grouping, } \\ & &\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\text{two middle terms are needed that multiply to a product of \(−24\) and add to a sum of \(−2\).} \\ & &\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; \text{\(6\) and \(4\) are good candidates; Since the product must be negative, one of these values must be negative.} \\ & &\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; \text{\(−6\) and positive \(4\) will work, because their product is \(−24\) and their sum is \(−2\).} \\ &8x^2 + 4x − 6x − 3 &\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\text{Four terms, the sum of the two middle terms is the original middle term, \(−2x\).} \\ & &\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; \text{The order of the two middle terms does not matter. } \\ &(8x^2 + 4x) + (−6x − 3) &\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\text{Create pairs of terms. Notice the addition in between the parentheses;} \\ & &\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\text{the third of the four terms was negative here, so the sign stays with the term.} \\ &4x(2x + 1) + (−3)(2x + 1) &\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\text{Factor out the GCF from each pair- a repeated binomial factor is always present} \\ &(4x − 3)(2x + 1) &\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\text{Solution. Be sure to check by FOIL.} \end{array}\)
    1. \(\begin{array} && 12x − 14x^3 + 22x^2 &\text{Example problem} \\ &−14x^3 + 22x^2 + 12x &\text{Reorder the terms in decreasing order of variable degree.} \\ &2x(−7x^2 + 11x + 6) &\text{Factor out the GCF so a trinomial results that can be factored using factor by grouping.} \\ & &\text{The GCF of \(2x\) will be included in the final answer, so do not forget about it.} \\ &−7x^2 + 11x + 6 &\text{Product \(ac\) is \(−7 ∗ 6 = −42\), Sum is \(b = 11\). To use factor by grouping, } \\ & &\text{two middle terms are needed that multiply to a product of \(−42\) and add to a sum of \(11\).} \\ & &\text{There are no numbers that fulfill both of these requirements,} \\& &\text{which means that the trinomial is not factorable into integer factors.} \\ &−7x^2 + 11x + 6 &\text{To find the factors and the zeroes of the polynomial, use the Quadratic Formula.} \\ & &\text{Let \(a = −7\), \(b = 11\), \(c = 6\)} \\ &x = \dfrac{−11 ± \sqrt{11^2 − 4(−7)(6)}}{2(−7)} &\text{Quadratic Formula} \\ &x = \dfrac{−11 ± \sqrt{121 + 168}}{-14} &\text{Simplify} \\ &x = \dfrac{11 ± \sqrt{289}}{14} &\text{Divide out \(−1\) from all terms} \\ &x = \dfrac{11 ± \sqrt{289}}{14} = 2, \;\;x = \dfrac{11 − \sqrt{289}}{14} = − \dfrac{3}{7} &\text{Exact answers for the zeroes in radical form, followed by real number form.} \\ &(x − 2),\;\; (x + -\dfrac{3}{7} ) &\text{Factors. Take care to insert the correct \(±\) in the factors.} \\ & & \text{Find the solutions, and then reverse-engineer to figure out the factor that will produce that solution.} \\ & & \text{The first solution from the Quadratic Formula was \(x = 2\).} \\ & & \text{A factor of \((x − 2)\) when set equal to \(0\) will produce the solution of \(x = 2\).} \\ & & \text{The second solution from the Quadratic Formula was \(x = −\dfrac{3}{7}\).} \\ & & \text{A factor of \((x + −\dfrac{3}{7})\) will produce the solution of \(x = −\dfrac{3}{7}\).} \\ &2x(x − 2)(x + \dfrac{3}{7} ) &\text{Polynomial factors, including the original GCF that was factored out at the beginning of this problem.} \end{array}\)
    1. \(\begin{array} && \dfrac{(x^2 + 1)^2 (−2) + (2x)2(x^2 + 1)(2x)}{(x^2 + 1)^4} &\;\;\;\;\;\;\;\;\;\;\;\;\text{Example problem} \\ &\dfrac{2(x^2 + 1)[(x^2 + 1)(−1) + (x)2(2x)]}{(x^2 + 1)(x^2 + 1)^3} &\;\;\;\;\;\;\;\;\;\;\;\;\text{Factor out the GCF from the numerator.} \\ &\dfrac{2\cancel{(x^2 + 1)}[(x^2 + 1)(−1) + (x)2(2x)]}{\cancel{(x^2 + 1)}(x^2 + 1)^3} &\;\;\;\;\;\;\;\;\;\;\;\;\text{Remove common factors.} \\ &\dfrac{2[(x^2 + 1)(−1) + (x)2(2x)]}{(x^2 + 1)^3} &\;\;\;\;\;\;\;\;\;\;\;\;\text{Remove common factors.} \\ &\dfrac{2[−x^2 − 1 + 4x^2]}{(x^2 + 1)^3} &\;\;\;\;\;\;\;\;\;\;\;\;\text{Simplify} \\ &\dfrac{2(3x^2 − 1)}{(x^2 + 1)^3} &\;\;\;\;\;\;\;\;\;\;\;\;\text{Final answer.} \end{array}\)
    1. \(\begin{array} &&\dfrac{(2x + 1)^{\frac{1}{2}} − (x + 2)(2x + 1)^{-\frac{1}{2}}}{2x + 1} &\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\text{Example problem} \\ &\dfrac{(2x + 1)^{\frac{1}{2}} −\dfrac{(x + 2)}{(2x + 1)^{\frac{1}{2}}}}{2x + 1} &\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\text{Write the expression with a positive exponent (move it to the denominator).} \\ &\dfrac{(2x + 1)^{\frac{1}{2}}(2x + 1)^{\frac{1}{2}} − (x + 2)}{\dfrac{(2x + 1)^{\frac{1}{2}}}{2x + 1}} &\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\text{Write the numerator with a common denominator.} \\ &\dfrac{2x + 1 − x − 2}{(2x + 1)^{\frac{1}{2}} (2x + 1)} &\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\text{Simplified.} \\ &\dfrac{2x + 1 − x − 2}{(2x + 1)^{\frac{3}{2}}} &\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\text{Final answer.} \end{array}\)
    Exercise 8.3.1

    Factor using any method discussed in this section:

    1. \(5x^2 − 23x − 10\)
    2. \(8x^2 + 2x − 3\)
    3. \(3x^2 − 7x − 6\)
    4. \(10x^2 + 13x − 5\)
    5. \(12x^5 − 17x^4 + 6x^3\)
    6. \(\dfrac{(2x^2 − 1)^2 (−2) + (2x)2(2x^2 − 1)(2x)}{(2x^2 − 1)^4}\)
    7. \(\dfrac{2(2x − 3)^{\frac{1}{3}} − (x − 1)(2x − 3)^{-\frac{2}{3}}}{2x − 3^{\frac{2}{3}}}\)

    Quadratic Formula

    Definition: Quadratic Formula

    The Quadratic Formula is used to solve (or find the zeroes) of a Polynomial (Quadratic Equation) of degree \(2\) that is in the form \(ax^2 + bx + c = 0\). The Quadratic Formula is:

    \[x = \dfrac{−b ± \sqrt{b^2 − 4ac}}{2a} \nonumber \]

    where \(a\), \(b\), and \(c\) are the coefficients of the standard form of a Quadratic Equation, \(ax^2 + bx + c = 0\).

    Example 8.3.2

    For the following functions, find all zeros of \(f\) using the Quadratic Formula. Express the final answer as exact answers (in radical form) and also as decimals, rounded to the thousandths place.

    1. \(f(x) = −2x^2 + 4x − 1\)
    2. \(f(x) = x^3 − 3x^2 − 4x\)
    Solution
    1. Set \(f(x) = 0: −2x^2 + 4x − 1 = 0\). This function is written in the form \(ax^2 + bx + c = 0\), with \(a = −2\), \(b = 4\) and \(c = −1\).

    Replacing \(a\), \(b\) and \(c\) in the Quadratic Formula with these values:

    \(\begin{array} &&x = \dfrac{−4 ± \sqrt{4^2 − 4(−2)(−1)}}{2(−2)} &\;\;\;\;\;\text{Quadratic Formula} \\ &x = \dfrac{−4 ± \sqrt{(16 − 8)}}{−4} &\;\;\;\;\;\text{Simplify} \\ &x = \dfrac{−4 ± \sqrt{8}}{−4} &\;\;\;\;\;\text{Simplify} \\ &x = \dfrac{−4 ± 2 \sqrt{2}}{−4} &\;\;\;\;\;\text{Simplify the radical} \\ &x = \dfrac{2 ± \sqrt{2}}{2} &\;\;\;\;\;\text{Exact answers in radical form} \\ &x = \dfrac{2 − \sqrt{2}}{2} ,\;\; x = \dfrac{2 + \sqrt{2}}{2} &\;\;\;\;\;\text{Exact answers written as two roots} \\ &x = 0.293 \text{ and } x = 1.707 &\;\;\;\;\;\text{Approximation answers rounded to the thousandths place} \end{array}\)

    1. The function \(f(x) = x^3 − 3x^2 − 4x\) is a cubic function. Factor out \(x\) from all three terms before using the Quadratic Equation on the trinomial factor: \(x(x^2 − 3x − 4) = 0\), with \(a = 1\), \(b = −3\) and \(c = −4\).

    Don’t forget that the \(x\) that was factored out is a root, namely \(x = 0\).

    Replacing \(a\), \(b\) and \(c\) in the Quadratic Formula with these values:

    \(\begin{array} &&x = \dfrac{3 ± \sqrt{(−3)2 − 4(1)(−4)}}{2(1)} &\text{Quadratic Formula} \\ &x = \dfrac{3 ± \sqrt{(16 + 9)}}{2} &\text{Simplify} \\ &x = \dfrac{3 ± \sqrt{25}}{2} &\text{Simplify} \\ &x = \dfrac{3 ± 5}{2} &\text{Simplify further} \\ &x = \dfrac{3 − 5}{2} ,\;\;x = \dfrac{−2}{2} ,\;\; x = −1 &\text{Second root (first root is \(x = 0\))} \\ &x = \dfrac{3 + 5}{2} , \;\;x = \dfrac{8}{2} , \;\;x = 4 &\text{Third root} \end{array}\)

    There are three solutions, or roots of the cubic function \(f(x) = x^3 − 3x^2 − 4x: x = 0\), \(x = −1\) and \(x = 4\).

    Exercise 8.3.2

    For the following functions, find all zeros of \(f\) using the Quadratic Formula. Express the final answer as exact answers (in radical form) and also as decimals, rounded to the thousandths place.

    1. \(f(t) = 9t^3 − 18t^2 + 6t\)
    2. \(f(x) = x^5 − 4x^4 − 32x^3\)
    3. \(f(x) = 18 − 3x − 2x^2\)
    4. \(f(x) = 12x^2 + 11x − 5\)
    5. \(f(x) = 3x^2 − 6x + 2\)

    This page titled 8.3: Factoring and Finding Polynomial Solutions (Zeroes) is shared under a CC BY-SA 4.0 license and was authored, remixed, and/or curated by Victoria Dominguez, Cristian Martinez, & Sanaa Saykali (ASCCC Open Educational Resources Initiative) .