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Mathematics LibreTexts

1.1: Solve Polynomial Equations by Factoring

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    38236
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    Reviewing General Factoring Strategies

    We have learned various techniques for factoring polynomials with up to four terms. The challenge is to identify the type of polynomial and then decide which method to apply. The following outlines a general guideline for factoring polynomials.

    General guidelines for factoring polynomials

    Step 1: Check for common factors. If the terms have common factors, then factor out the greatest common factor (GCF).

    Step 2: Determine the number of terms in the polynomial.

    1. Factor four-term polynomials by grouping (either GCF of pairs, or binomial square then difference of squares).
    2. Factor trinomials (3 terms) using “trial and error” or the AC method.
      • Possibly a Binomial Square, which has the form: \(a^{2}+2ab+b^{2}=(a+b)^2\) or \(a^{2}-2ab+b^{2}=(a-b)^2\)
    3. Factor binomials (2 terms) using the following special products:
      • Difference of squares:\(a^{2}−b^{2}=(a−b)(a+b)\)
      • Sum of squares: \(a^{2}+b^{2}\)  no general formula
      • Difference of cubes: \(a^{3}−b^{3}=(a−b)(a^{2}+ab+b^{2})\)
      • Sum of cubes: \(a^{3}+b^{3}=(a+b)(a^{2}−ab+b^{2})\)
      • If a binomial is both a difference of squares and a difference cubes, then first factor it as difference of squares. This will result in a more complete factorization.

    Step 3: Look for factors that can be factored further.

    Step 4: Check by multiplying.

    Use Factoring to Solve Equations

    We will first solve some equations by using the Zero Factor Property. The Zero Factor Property (also called the Zero Product Property) says that if the product of two quantities is zero, then at least one of those quantities is zero. The only way to get a product equal to zero is to multiply by zero itself.

    Zero Factor Property

    If \(a·b=0\), then either \(a=0\) or \(b=0\) or both.

    For example, consider the equation \( (x - 3)(x - 2) = 0 \). According to the Zero Factor Property, this product can only be zero if one of the factors is zero. For this equation, the factors are \( (x-3) \) and \( (x-2) \). Factors are the expressions that are multiplied together to form a \( \underline{\text{product}} \).

    \( (x - 3)(x - 2) = 0 \)
    \( \begin{align*} x-3 &=0  & \text{or}  &&   x-2 & =0 \\
    x &=3 &&& x &=2 
    \end{align*} \)
    \( \text{Solution Set: } \{3, 2\} \)

    These proposed solutions can be checked by substituting back in the original equation.

    \( \begin{align*}
    \text{ Check } x &   ={\color{Cerulean}{3}}     & \text{and}  &&   \text{ Check } x &={\color{Cerulean}{2}}  \\
    ({\color{Cerulean}{3}} - 3)({\color{Cerulean}{3}} - 2) &=0  &&&   ({\color{Cerulean}{2}} - 3)({\color{Cerulean}{2}} - 2) & =0 \\
    (0)(1) &=0  \quad {\color{Cerulean}{✓}} &&& (-1)(0) &=0 \quad {\color{Cerulean}{✓}} 
    \end{align*} \)

    how-to.pngHow to: Use the Zero Factor Property to Solve an Equation.

    1. ZERO. Write the equation so one side of the equation is  zero. Write the expression on the other side of the equal sign in order of descending powers of \(x\) with a positive coefficient on the term with highest exponent.
    2. FACTORFactor the expression.
    3. PROPERTY. Set each factor equal to zero and solve. (This is the property -- a factor that is zero will make the product of factors it is a part of also equal to zero).  The solutions obtained are the values of \(x\) that will make the original equation a true statement.  
    4. Check by substituting solutions into the original equation.

    Example \(\PageIndex{1}\) Factor out a GCF

    Solve: \(2 x ^2  = 8x \).

    Solution.  Notice that the first step requires one side of the equation to be made zero. If both sides of the equal sign instead were first divided by x, then only one solution \( x=4\) would have been found. Dividing by a variable expression can result in lost solutions!

    \(\begin{align*}
        &  &2x^2-8x&=0  &&  && \text{1. } \underline{\textbf{Zero}} \text{. Make one side zero} \\
        &  &2x(x-4)&= 0  &&  &&  \text{2. }  \underline{\textbf{Factor}} \text{. Factor out the GCF}\\
    2x &=0  &\text{or}&  &x-4& =0  && \text{3. } \underline{\textbf{Property}} \text{. Set each factor to 0}  \\
    x&=0  &&  &x&=4  && \quad \text{And solve} \\
     &  & \text{Solution Set: }&\{  0, 4 \}  \end{align*}\)

    Example \(\PageIndex{2}\) Factor four terms by grouping pairs

    Solve: \(4 x ^ { 3 } - x ^ { 2 } - 100 x + 25 = 0\).

    Solution

    \(\begin{align*}
        &  &4 x ^ { 3 } - x ^ { 2 } - 100 x + 25  = 0&  &&  &&  \text{1. } \underline{\textbf{Zero}} \text{. One side already zero} \\
        &  &x ^ { 2 } ( 4 x - 1 ) - 25 ( 4 x - 1 )=0&  &&  &&  \text{2. }  \underline{\textbf{Factor}} \text{. Factor by grouping pairs}\\
        &  &( 4 x - 1 ) \left( x ^ { 2 } - 25 \right)=0&  &&  && \quad\text{Factor out the common binomial}\\
        &  &( 4 x - 1 ) ( x - 5 ) ( x + 5 )=0&  &&  && \quad\text{Factor a difference of squares}\\
    4x-1&=0  &x-5=0\quad\qquad&  &x+5&=0  && \text{3. } \underline{\textbf{Property}} \text{. Set each factor to 0}  \\
    4x&=1  &x=5\qquad\qquad&  &x&=-5  && \quad \text{And solve} \\
    x&=\frac{1}{4}  &&  &&  && \\
     &  & \text{Solution Set: } \Big\{ \frac{1}{4}, 5, -5 \Big\}&  \end{align*}\) 

    Example \(\PageIndex{3}\) Factor a trinomial (with  a constant GCF and then \(a = 1\))

    Solve: \(3x^2=12x+63\).

    Solution.  This example  highlights the essential first step of making one side of the equation zero before the Zero Factor Property is applied.  Also, factoring produces three factors, but the first factor is a constant \( (3) \) which can never be equal to \( 0 \)

    \(\begin{align*}
        &  &3x^2−12x−63=0&  &&  && \text{1. } \underline{\textbf{Zero}} \text{. Make one side zero} \\
        &  &3(x^2−4x−21)=0&  &&  &&  \text{2. }  \underline{\textbf{Factor}} \text{. Factor out the GCF}\\
        &  &3(x−7)(x+3)=0&  &&  && \quad\text{Factor the trinomial}\\
    3&=0  &x-7=0\quad&  &x-5&=0  && \text{3. } \underline{\textbf{Property}} \text{. Set each factor to 0}  \\
    3&\neq 0  &x=7\qquad&  &x&=5  && \quad \text{And solve} \\
     &  & \text{Solution Set: }\{  7, 5 \}&  \end{align*}\)

    Example \(\PageIndex{4}\) Factor a trinomial (with \(a \ne 1\))

    Solve: \(15 x ^ { 2 } + 3 x - 8 = 5 x - 7\).

    Solution

    \( \begin{array}{cl}
      15 x ^ { 2 } + 3 x - 8 = 5 x - 7 & \\
      15 x ^ { 2 } - 2 x - 1 = 0   & \qquad \text{1. } \underline{\textbf{Zero}} \text{. Make one side zero}\\
      (3x−1)(5x+1)=0  & \qquad  \text{2. }  \underline{\textbf{Factor}} \text{. Factor the trinomial}\\
     \begin{array} {ccc} 3x−1=0 & \text{ and }  & 5x+1=0 \\
    3x = 1 & & 5x = -1 \\
    x = \frac { 1 } { 3 } & & x = -\frac { 1 } { 5 } \end{array}
    & \begin{array} {l}  \qquad  \text{3. } \underline{\textbf{Property}} \text{. Set each factor to 0}\\
     \quad \qquad \text{And solve} \\  \text{  } \end{array} \\
     \text{Solution Set: }  \Big\{  \frac { 1 } { 3 }, -\frac { 1 } { 5 } \Big\}  
    \end{array} \)

    Example \(\PageIndex{5}\) Factor a trinomial (with \(a \ne 1\))

    Solve: \((3x−8)(x−1)=3x\).

    Solution. This quadratic equation appears to be factored; hence it might be tempting to set each factor equal to \(3x\). However, this would lead to incorrect results. We must first rewrite the equation equal to zero, so that we can apply the zero-product property

    \( \begin{array}{cccl}
     & (3x−8)(x−1)=3x & \\
     & 3x^2−11x+8=3x   &  & \qquad \text{1. } \underline{\textbf{Zero}} \text{. Make one side zero}\\
     & 3x^2−14x+8=0   &  &  \\
     & (3x−2)(x−4)=0   & & \qquad  \text{2. }  \underline{\textbf{Factor}} \text{. Factor the trinomial}\\
    3x−2=0 & \text{ and }  & x-4=0 & \qquad  \text{3. } \underline{\textbf{Property}} \text{. Set each factor to 0}\\
    3x = 2 & & x=4 & \quad \qquad \text{And solve} \\
    x = \frac { 2 } { 3 } & & \\
     &\text{Solution Set: }  \Big\{  \frac { 2 } { 3 }, 4 \Big\}  &
    \end{array} \)

    Example \(\PageIndex{6}\) Factor a difference of squares

    Solve: \(169q^2=49\).

    Solution

    \( \begin{array}{cccl}
     & 169q^2=49 & \\
     & 169x^2−49=0   &  & \qquad \text{1. } \underline{\textbf{Zero}} \text{. Make one side zero}\\
     & (13x−7)(13x+7)=0  & & \qquad  \text{2. }  \underline{\textbf{Factor}} \text{. Factor a difference of squares}\\
    13x−7=0 & \text{ and }  & 13x+7=0 & \qquad  \text{3. } \underline{\textbf{Property}} \text{. Set each factor to 0}\\
    13x = 7 & & 13x=-7 & \quad \qquad \text{And solve} \\
    x = \frac {7 } { 13 } & & x = -\frac {7 } { 13 }\\
     &\text{Solution Set: }  \Big\{  \frac {7 } { 13 }, -\frac {7 } { 13 } \Big\}  &
    \end{array} \)

     

    Example \(\PageIndex{7}\) Factor a perfect square trinomial (with a variable GCF)

    Solve: \(9m^3+100m=60m^2\)

    Solution

    \(\begin{array}{cl}
         9m^3+100m=60m^2 &  \\
       9m^3−60m^2+100m=0      & \text{1. } \underline{\textbf{Zero}} \text{. Make one side zero} \\
        m(9m^2−60m+100)=0        &  \text{2. }  \underline{\textbf{Factor}} \text{. Factor out the GCF}\\
       m(3m−10)^2=0    &      \text{Factor the Perfect Square trinomial}\\
     \begin{array} {ccc}
    m=0  & \text{ and } & 3m-10=0 \\
    & &  m=10 \\
    & & m=\frac{10}{3}
    \end{array}

    \begin{array} {l}  \text{3. } \underline{\textbf{Property}} \text{. Set each factor to 0}\\
       \text{And solve} \\  \text{  } \end{array}
    \\
    \text{Solution Set: } \Big\{  0, \frac{10}{3} \Big\}  \end{array}\)

    Example \(\PageIndex{8}\) Factor four terms (the difference of a binomial square and a monomial square)

    Solve: \(4x^{2} + 36x + 81 = 100 x^2\).

    Solution

    \( \begin{array}{cl}
      4x^2 + 36x + 81 = 100 x^2 & \\
     4x^2 + 36x + 81 − 100 x^2 = 0   & \qquad \text{1. } \underline{\textbf{Zero}} \text{. Make one side zero}\\
      {\color{Cerulean}{4x^2}} +36x +  {\color{Cerulean}{81}}  − {\color{Red}{100 x^2}}  & \qquad \quad \text{Observe three terms are perfect squares (2 positive, 1 negative)}\\
    (2x+9)(2x+9)− 100 x^2=0  & \qquad  \text{2a. }  \underline{\textbf{Factor}} \text{. Factor the perfect square binomial}\\
    (2x+9)^2− (10x)^2=0  &\\
    (2x+9 − 10x)(2x+9 + 10x)=0  & \qquad  \text{2b. }  \underline{\textbf{Factor}} \text{. Factor the difference of squares}\\
     \begin{array} {ccc} -8x+9=0 & \text{ and }  & 12x+9=0 \\
    8x = 9 & & 12x = -9 \\
    x = \dfrac { 9 } { 8 } & & x = -\dfrac { 9 } { 12 } \end{array}
    & \begin{array} {l}  \qquad  \text{3. } \underline{\textbf{Property}} \text{. Set each factor to 0}\\
     \quad \qquad \text{And solve} \\  \text{  } \end{array} \\
     \text{Solution Set: }  \Big\{  \dfrac { 9 } { 8 }, -\dfrac { 3 } { 4 } \Big\}  
    \end{array} \)

    Factoring using the sum or difference of cubes formula to solve an equation will be discussed in the next section that includes the Quadratic Formula and the Complete the Square technique. 

    try-it.png Try It \(\PageIndex{9}\)

    Solve each equation by factoring.

    1.  \((3m−2)(2m+1)=0\)
    2.  \( 3c^2=10c−8\)
    3.  \(25p^2=49\)
    1. \((2m+1)(m+3)=12m\)
    2.  \(123b=−6−60b^2\)
    3.  \(8x^3=24x^2−18x\)
    1. \( 64x^2 + 225 = 240x \)
    2. \(3x^3 - 2x^2 - 12x + 8 = 0 \)
    3. \( 9x^2 + 16 =27x +  64 \)
    Answer

    a. \(m=\frac{2}{3},\space m=−\frac{1}{2}\)

    b. \( c=2,\space c=\frac{4}{3}\)

    c. \(p=\frac{7}{5},p=−\frac{7}{5}\)

    d. \(m=1,\space m=\frac{3}{2}\)

    e. \(b=−2,\space b=−\frac{1}{20}\)

    f. \(x=0,\space x=\frac{3}{2}\)

    g. \(x=\dfrac{15}{8}\)

    h. \(x=2,\space x=-2,\space x=\dfrac{2}{3}\)

    i. \(x=\dfrac{4}{3}\space x=-\dfrac{9}{4}\)


    1.1: Solve Polynomial Equations by Factoring is shared under a not declared license and was authored, remixed, and/or curated by LibreTexts.

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