# 6.8: Factoring Polynomials in Quadratic Form

**At Grade**Created by: CK-12

**Practice**Methods for Solving Quadratic Functions

The volume of a rectangular prism is \begin{align*}10x^3 - 25x^2 - 15x\end{align*}. 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*}. Another possibility is something similar to the difference of squares, \begin{align*}a^4-b^4\end{align*}. This can be factored to \begin{align*}(a^2-b^2)(a^2+b^2)\end{align*} or \begin{align*}(a-b)(a+b)(a^2+b^2)\end{align*}. 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*}.

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

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

#### Example B

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

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

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

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

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

\begin{align*}9x^2+4\end{align*} 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*}.

**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*}

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

Factor \begin{align*}x^2-9\end{align*} further and solve for \begin{align*}x\end{align*} where possible. \begin{align*}2x^2+1\end{align*} 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*}

**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*}.

First, pull out the GCF among the three terms.

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

Factor what is inside the parenthesis like a quadratic equation. \begin{align*}ac = -6\end{align*} 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*}

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

### Guided Practice

Factor the following polynomials.

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

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

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

#### Answers

1. \begin{align*}ac = 24\end{align*} 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*}

2. Factor this polynomial like a difference of squares.

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

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

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

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*}

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.

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

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

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

### Notes/Highlights Having trouble? Report an issue.

Color | Highlighted Text | Notes | |
---|---|---|---|

Please Sign In to create your own Highlights / Notes | |||

Show More |

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 that make each binomial equal to zero.factored form

The factored form of a quadratic function is , where and 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 , where , , and are real constants and .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 .Vertex form

The vertex form of a quadratic function is , where is the vertex of the parabola.Zeroes of a Polynomial

The zeroes of a polynomial are the values of that cause to be equal to zero.### Image Attributions

Here you'll how to factor and solve polynomials that are in “quadratic form.”