Skip to main content
K12 LibreTexts

2.1.1: Methods for Solving Quadratic Functions

  • Page ID
    14122
  • Factoring Polynomials in Quadratic Form

    The volume of a rectangular prism is 10x3−25x2−15x. What are the lengths of the prism's sides?


    Factoring Polynomials in Quadratic Form 

    The last type of factorable polynomial are those that are in quadratic form. Quadratic form is when a polynomial looks like a trinomial or binomial and can be factored like a quadratic. One example is when a polynomial is in the form ax4+bx2+c. Another possibility is something similar to the difference of squares, a4−b4. This can be factored to (a2−b2)(a2+b2) or (a−b)(a+b)(a2+b2). Always keep in mind that the greatest common factors should be factored out first.

    Let's factor the following polynomials.

    1. 2x4−x2−15

      This particular polynomial is factorable. First, ac=−30. The factors of -30 that add up to -1 are -6 and 5. Expand the middle term and then use factoring by grouping.

      2x4−x2−15

      2x4−6x2+5x2−15

      2x2(x2−3)+5(x2−3)

      (x2−3)(2x2+5)

      Both of the factors are not factorable, so we are done.

    2. 81x4−16

      Treat this polynomial equation like a difference of squares.

      81x4−16

      (9x2−4)(9x2+4)

      Now, we can factor 9x2−4 using the difference of squares a second time.

      (3x−2)(3x+2)(9x2+4)

      9x2+4 cannot be factored because it is a sum of squares. This will have imaginary solutions.

      Now, let's find all the real-number solutions of 6x5−51x3−27x=0.

      First, pull out the GCF among the three terms.

      6x5−51x3−27x=0

      3x(2x4−17x2−9)=0

      Factor what is inside the parenthesis like a quadratic equation. ac=−18 and the factors of -18 that add up to -17 are -18 and 1. Expand the middle term and then use factoring by grouping.

      6x5−51x3−27x=0

      3x(2x4−17x2−9)=0

      3x(2x4−18x2+x2−9)=0

      3x[2x2(x2−9)+1(x2−9)]=0

      3x(x2−9)(2x2+1)=0

      Factor x2−9 further and solve for x where possible. 2x2+1 is not factorable.

      3x(x2−9)(2x2+1)=0

      3x(x−3)(x+3)(2x2+1)=0

      x=−3,0,3


    Examples

    Example 1

    Earlier, you were asked to find the lengths of the prism's sides. 

    Solution

    To find the lengths of the prism's sides, we need to factor 10x3−25x2−15x.

    First, pull out the GCF among the three terms.

    10x3−25x2−15x

    5x(2x2−5x−3)

    Factor what is inside the parenthesis like a quadratic equation. ac=−6 and the factors of -6 that add up to -5 are -6 and 1.

    5x(2x2−5x−3)=5x(2x+1)(x−3)

    Therefore, the lengths of the rectangular prism's sides are 5x, 2x+1, and x−3.

    Example 2

    Factor: 3x4+14x2+8.

    Solution

    ac=24 and the factors of 24 that add up to 14 are 12 and 2.

    3x4+14x2+8

    3x4+12x2+2x2+8

    3x2(x2+4)+2(x4+4)

    (x2+4)(3x2+2)

    Example 3

    Factor: 36x4−25.

    Solution

    Factor this polynomial like a difference of squares.

    36x4−25

    (6x2−5)(6x2+5)

    6 and 5 are not square numbers, so this cannot be factored further.

    Example 4

    Find all the real-number solutions of 8x5+26x3−24x=0.

    Solution

    Pull out a 2x from each term.

    8x5+26x3−24x=0

    2x(4x4+13x−12)=0

    2x(4x4+16x2−3x2−12)=0

    2x[4x2(x2+4)−3(x2+4)]=0

    2x(x2+4)(4x2−3)=0

    Set each factor equal to zero.

    4x2−3=0

    2x=0

    x2+4=0

    and x2= \(\ 3 \over 4\)

    x=0

    x2=−4

    x=± \(\ \frac{\sqrt{3}}{2}\)

    Notice the second factor will give imaginary solutions.


    Review

    Factor the following quadratics completely.

    1. x4−6x2+8
    2. x4−4x2−45
    3. x4−18x2+45
    4. 4x4−11x2−3
    5. 6x4+19x2+8
    6. x4−81
    7. 16x4−1
    8. 6x5+26x3−20x
    9. 4x6−36x2
    10. 625−81x4

    Find all the real-number solutions to the polynomials below.

    1. 2x4−5x2−12=0
    2. x4−16=0
    3. 16x4−49=0
    4. 12x6+69x4+45x2=0
    5. 3x4+17x2−6=0

    Answers for Review Problems

    To see the Review answers, open this PDF file and look for section 6.8. 


    Vocabulary

    Term Definition
    Factor to Solve "Factor to Solve" is a common method for solving quadratic equations accomplished by factoring a trinomial into two binomials and identifying the values of x that make each binomial equal to zero.
    factored form The factored form of a quadratic function f(x) is f(x)=a(x−r1)(x−r2), where r1 and r2 are the roots of the function.
    Factoring Factoring is the process of dividing a number or expression into a product of smaller numbers or expressions.
    Quadratic form A polynomial in quadratic form looks like a trinomial or binomial and can be factored like a quadratic expression.
    quadratic function A quadratic function is a function that can be written in the form f(x)=ax2+bx+c, where a, b, and c are real constants and a≠0.
    Roots The roots of a function are the values of x that make y equal to zero.
    standard form The standard form of a quadratic function is f(x)=ax2+bx+c.
    Vertex form The vertex form of a quadratic function is y=a(x−h)2+k, where (h,k) is the vertex of the parabola.
    Zeroes of a Polynomial The zeroes of a polynomial f(x) are the values of x that cause f(x) to be equal to zero.
    • Was this article helpful?