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6.8: Factoring Polynomials in Quadratic Form

Difficulty Level: At Grade Created by: CK-12
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The volume of a rectangular prism is \begin{align*}10x^3 - 25x^2 - 15x\end{align*}10x325x215x. What are the lengths of the prism's sides?

Guidance

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 \begin{align*}ax^4+bx^2+c\end{align*}ax4+bx2+c. Another possibility is something similar to the difference of squares, \begin{align*}a^4-b^4\end{align*}a4b4. This can be factored to \begin{align*}(a^2-b^2)(a^2+b^2)\end{align*}(a2b2)(a2+b2) or \begin{align*}(a-b)(a+b)(a^2+b^2)\end{align*}(ab)(a+b)(a2+b2). Always keep in mind that the greatest common factors should be factored out first.

Example A

Factor \begin{align*}2x^4-x^2-15\end{align*}2x4x215.

Solution: This particular polynomial is factorable. Let’s use the method we learned in the Factoring when \begin{align*}a \neq 1\end{align*}a1 concept. First, \begin{align*}ac = -30\end{align*}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.

\begin{align*}& 2x^4-x^2-15\\ & 2x^4-6x^2+5x^2-15\\ & 2x^2(x^2-3)+5(x^2-3)\\ & (x^2-3)(2x^2+5)\end{align*}2x4x2152x46x2+5x2152x2(x23)+5(x23)(x23)(2x2+5)

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

Example B

Factor \begin{align*}81x^4-16\end{align*}81x416.

Solution: Treat this polynomial equation like a difference of squares.

\begin{align*}& 81x^4-16\\ & (9x^2-4)(9x^2+4)\end{align*}81x416(9x24)(9x2+4)

Now, we can factor \begin{align*}9x^2-4\end{align*}9x24 using the difference of squares a second time.

\begin{align*}(3x-2)(3x+2)(9x^2+4)\end{align*}(3x2)(3x+2)(9x2+4)

\begin{align*}9x^2+4\end{align*}9x2+4 cannot be factored because it is a sum of squares. This will have imaginary solutions.

Example C

Find all the real-number solutions of \begin{align*}6x^5-51x^3-27x = 0\end{align*}6x551x327x=0.

Solution: First, pull out the GCF among the three terms.

\begin{align*}6x^5-51x^3-27x &= 0\\ 3x(2x^4-17x^2-9) &= 0\end{align*}6x551x327x3x(2x417x29)=0=0

Factor what is inside the parenthesis like a quadratic equation. \begin{align*}ac = -18\end{align*}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.

\begin{align*}6x^5-51x^3-27x &= 0\\ 3x(2x^4-17x^2-9) &= 0\\ 3x(2x^4-18x^2+x^2-9) &= 0\\ 3x[2x^2(x^2-9)+1(x^2-9)] &= 0\\ 3x(x^2-9)(2x^2+1) &= 0\end{align*}6x551x327x3x(2x417x29)3x(2x418x2+x29)3x[2x2(x29)+1(x29)]3x(x29)(2x2+1)=0=0=0=0=0

Factor \begin{align*}x^2-9\end{align*}x29 further and solve for \begin{align*}x\end{align*}x where possible. \begin{align*}2x^2+1\end{align*}2x2+1 is not factorable.

\begin{align*}3x(x^2-9)(2x^2+1) &= 0\\ 3x(x-3)(x+3)(2x^2+1) &= 0\\ x &= -3, 0, 3\end{align*}3x(x29)(2x2+1)3x(x3)(x+3)(2x2+1)x=0=0=3,0,3

Intro Problem Revisit To find the lengths of the prism's sides, we need to factor \begin{align*}10x^3 - 25x^2 - 15x\end{align*}10x325x215x.

First, pull out the GCF among the three terms.

\begin{align*}10x^3 - 25x^2 - 15x\\ 5x(2x^2 - 5x - 3)\end{align*}10x325x215x5x(2x25x3)

Factor what is inside the parenthesis like a quadratic equation. \begin{align*}ac = -6\end{align*}ac=6 and the factors of -6 that add up to -5 are -6 and 1.

\begin{align*}5x(2x^2 - 5x - 3) = 5x(2x + 1)(x - 3)\end{align*}5x(2x25x3)=5x(2x+1)(x3)

Therefore, the lengths of the rectangular prism's sides are \begin{align*}5x\end{align*}5x, \begin{align*}2x + 1\end{align*}2x+1, and \begin{align*}x - 3\end{align*}x3.

Guided Practice

Factor the following polynomials.

1. \begin{align*}3x^4+14x^2+8\end{align*}3x4+14x2+8

2. \begin{align*}36x^4-25\end{align*}36x425

3. Find all the real-number solutions of \begin{align*}8x^5+26x^3-24x = 0\end{align*}8x5+26x324x=0.

Answers

1. \begin{align*}ac = 24\end{align*}ac=24 and the factors of 24 that add up to 14 are 12 and 2.

\begin{align*}& 3x^4+14x^2+8\\ & 3x^4+12x^2+2x^2+8\\ & 3x^2(x^2+4)+2(x^4+4)\\ & (x^2+4)(3x^2+2)\end{align*}3x4+14x2+83x4+12x2+2x2+83x2(x2+4)+2(x4+4)(x2+4)(3x2+2)

2. Factor this polynomial like a difference of squares.

\begin{align*}& 36x^4-25\\ & (6x^2-5)(6x^2+5)\end{align*}36x425(6x25)(6x2+5)

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

3. Pull out a \begin{align*}2x\end{align*}2x from each term.

\begin{align*}8x^5+26x^3-24x &= 0\\ 2x(4x^4+13x-12) &= 0\\ 2x(4x^4+16x^2-3x^2-12) &= 0\\ 2x[4x^2(x^2+4)-3(x^2+4)] &= 0\\ 2x(x^2+4)(4x^2-3) &= 0\end{align*}8x5+26x324x2x(4x4+13x12)2x(4x4+16x23x212)2x[4x2(x2+4)3(x2+4)]2x(x2+4)(4x23)=0=0=0=0=0

Set each factor equal to zero.

\begin{align*}& \qquad \qquad \qquad \qquad \qquad \qquad \ \ 4x^2-3 = 0\\ & 2x = 0 \quad x^2+4 = 0\\ & \qquad \quad \ \ \ \ \quad \qquad \qquad \ \ and \qquad \quad x^2 = \frac{3}{4}\\ & \ x = 0 \qquad \quad x^2 = -4\\ & \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad \ x = \pm \frac{\sqrt{3}}{2}\end{align*}  4x23=02x=0x2+4=0      andx2=34 x=0x2=4 x=±32

Notice the second factor will give imaginary solutions.

Vocabulary

Quadratic form
When a polynomial looks a trinomial or binomial and can be factored like a quadratic equation.

Practice

Factor the following quadratics completely.

  1. \begin{align*}x^4-6x^2+8\end{align*}
  2. \begin{align*}x^4-4x^2-45\end{align*}
  3. \begin{align*}x^4-9x^2+45\end{align*}
  4. \begin{align*}4x^4-11x^2-3\end{align*}
  5. \begin{align*}6x^4+19x^2+8\end{align*}
  6. \begin{align*}x^4-81\end{align*}
  7. \begin{align*}16x^4-1\end{align*}
  8. \begin{align*}6x^5+26x^3-20x\end{align*}
  9. \begin{align*}4x^6-36x^2\end{align*}
  10. \begin{align*}625-81x^4\end{align*}

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

  1. \begin{align*}2x^4-5x^2-12 = 0\end{align*}
  2. \begin{align*}x^4-16=0\end{align*}
  3. \begin{align*}16x^4-49 = 0\end{align*}
  4. \begin{align*}12x^6+69x^4+45x^2 = 0\end{align*}
  5. \begin{align*}3x^4+17x^2 -6=0\end{align*}

Vocabulary

Factor to Solve

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

factored form

The factored form of a quadratic function f(x) is f(x)=a(x-r_{1})(x-r_{2}), where r_{1} and r_{2} are the roots of the function.
Factoring

Factoring

Factoring is the process of dividing a number or expression into a product of smaller numbers or expressions.
Quadratic form

Quadratic form

A polynomial in quadratic form looks like a trinomial or binomial and can be factored like a quadratic expression.
quadratic function

quadratic function

A quadratic function is a function that can be written in the form f(x)=ax^2 + bx + c, where a, b, and c are real constants and a\ne 0.
Roots

Roots

The roots of a function are the values of x that make y equal to zero.
standard form

standard form

The standard form of a quadratic function is f(x)=ax^{2}+bx+c.
Vertex form

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

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.

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Date Created:
Mar 12, 2013
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Apr 28, 2016
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