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# Methods for Solving Quadratic Functions

## Factoring, completing the square, and the quadratic formula.

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Practice Methods for Solving Quadratic Functions
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The volume of a rectangular prism is $10x^3 - 25x^2 - 15x$ . 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 $ax^4+bx^2+c$ . Another possibility is something similar to the difference of squares, $a^4-b^4$ . This can be factored to $(a^2-b^2)(a^2+b^2)$ or $(a-b)(a+b)(a^2+b^2)$ . Always keep in mind that the greatest common factors should be factored out first.

#### Example A

Factor $2x^4-x^2-15$ .

Solution: This particular polynomial is factorable. Let’s use the method we learned in the Factoring when $a \neq 1$ concept. 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.

$& 2x^4-x^2-15\\& 2x^4-6x^2+5x^2-15\\& 2x^2(x^2-3)+5(x^2-3)\\& (x^2-3)(2x^2+5)$

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

#### Example B

Factor $81x^4-16$ .

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

$& 81x^4-16\\& (9x^2-4)(9x^2+4)$

Now, we can factor $9x^2-4$ using the difference of squares a second time.

$(3x-2)(3x+2)(9x^2+4)$

$9x^2+4$ cannot be factored because it is a sum of squares. This will have imaginary solutions.

#### Example C

Find all the real-number solutions of $6x^5-51x^3-27x = 0$ .

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

$6x^5-51x^3-27x &= 0\\3x(2x^4-17x^2-9) &= 0$

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.

$6x^5-51x^3-27x &= 0\\3x(2x^4-17x^2-9) &= 0\\3x(2x^4-18x^2+x^2-9) &= 0\\3x[2x^2(x^2-9)+1(x^2-9)] &= 0\\3x(x^2-9)(2x^2+1) &= 0$

Factor $x^2-9$ further and solve for $x$ where possible. $2x^2+1$ is not factorable.

$3x(x^2-9)(2x^2+1) &= 0\\3x(x-3)(x+3)(2x^2+1) &= 0\\x &= -3, 0, 3$

Intro Problem Revisit To find the lengths of the prism's sides, we need to factor $10x^3 - 25x^2 - 15x$ .

First, pull out the GCF among the three terms.

$10x^3 - 25x^2 - 15x\\5x(2x^2 - 5x - 3)$

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.

$5x(2x^2 - 5x - 3)= 5x(2x + 1)(x - 3)$

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

### Guided Practice

Factor the following polynomials.

1. $3x^4+14x^2+8$

2. $36x^4-25$

3. Find all the real-number solutions of $8x^5+26x^3-24x = 0$ .

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

$& 3x^4+14x^2+8\\& 3x^4+12x^2+2x^2+8\\& 3x^2(x^2+4)+2(x^4+4)\\& (x^2+4)(3x^2+2)$

2. Factor this polynomial like a difference of squares.

$& 36x^4-25\\& (6x^2-5)(6x^2+5)$

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

3. Pull out a $2x$ from each term.

$8x^5+26x^3-24x &= 0\\2x(4x^4+13x-12) &= 0\\2x(4x^4+16x^2-3x^2-12) &= 0\\2x[4x^2(x^2+4)-3(x^2+4)] &= 0\\2x(x^2+4)(4x^2-3) &= 0$

Set each factor equal to zero.

$& \qquad \qquad \qquad \qquad \qquad \qquad \ \ 4x^2-3 = 0\\& 2x = 0 \quad x^2+4 = 0\\& \qquad \quad \ \ \ \ \quad \qquad \qquad \ \ and \qquad \quad x^2 = \frac{3}{4}\\& \ x = 0 \qquad \quad x^2 = -4\\& \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad \ x = \pm \frac{\sqrt{3}}{2}$

Notice the second factor will give imaginary solutions.

### Vocabulary

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

### Practice

1. $x^4-6x^2+8$
2. $x^4-4x^2-45$
3. $x^4-18x^2+45$
4. $4x^4-11x^2-3$
5. $6x^4+19x^2+8$
6. $x^4-81$
7. $16x^4-1$
8. $6x^5+26x^3-20x$
9. $4x^6-36x^2$
10. $625-81x^4$

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

1. $2x^4-5x^2-12 = 0$
2. $x^4-16=0$
3. $16x^4-49 = 0$
4. $12x^6+69x^4+45x^2 = 0$
5. $3x^4+17x^2 -6=0$

### Vocabulary Language: English

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.

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

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.