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Difficulty Level: At Grade Created by: CK-12

In this lesson, we will learn how to factor quadratic polynomials for different values of a, b$a,\ b$, and c$c$. In the last lesson, we factored common monomials, so you already know how to factor quadratic polynomials where c=0$c=0$.

## Factoring Quadratic Expressions in Standard From

Quadratic polynomials are polynomials of degree 2. The standard form of a quadratic polynomial is ax2+bx+c$ax^2+bx+c$, where a, b$a,\ b$, and c$c$ are real numbers.

Example 1: Factor x2+5x+6$x^2+5x+6$.

Solution: We are looking for an answer that is a product of two binomials in parentheses: (x+ )(x+ )$(x+ \underline{\;\;\;\;\;\;\;\;\;\;\;} \ )(x + \underline{\;\;\;\;\;\;\;\;\;\;\;} \ )$.

To fill in the blanks, we want two numbers m$m$ and n$n$ that multiply to 6 and add to 5. A good strategy is to list the possible ways we can multiply two numbers to give us 6 and then see which of these pairs of numbers add to 5. The number six can be written as the product of.

So the answer is (x+2)(x+3)$(x+2)(x+3)$.

We can check to see if this is correct by multiplying (x+2)(x+3)$(x+2)(x+3)$.

x$x$ is multiplied by x$x$ and 3=x2+3x$3 =x^2+3x$.

2 is multiplied by x$x$ and 3=2x+6$3 =2x+6$.

Combine the like terms: x2+5x+6$x^2+5x+6$.

Example 2: Factor x26x+8$x^2-6x+8$.

Solution: We are looking for an answer that is a product of the two parentheses (x+ )(x+ )$(x+ \underline{\;\;\;\;\;\;\;\;\;\;\;} \ )(x + \underline{\;\;\;\;\;\;\;\;\;\;\;} \ )$.

The number 8 can be written as the product of the following numbers.

8=18$8=1\cdot 8$ and 1+8=9$1+8=9$ Notice that these are two different choices.

8&=(-1)(-8) && and && -1+(-8)=-9\\
8&=2 \times 4 && and && 2+4=6

And

8=(2)(4)$8=(-2)\cdot (-4)$ and 2+(4)=6$-2+(-4)=-6 \leftarrow$ This is the correct choice.

The answer is (x2)(x4)$(x-2)(x-4)$.

Example 3: Factor x2+2x15$x^2+2x-15$.

Solution: We are looking for an answer that is a product of two parentheses (x± )(x± )$(x \pm \underline{\;\;\;\;\;\;\;\;\;\;\;} \ )(x \pm \underline{\;\;\;\;\;\;\;\;\;\;\;} \ )$.

In this case, we must take the negative sign into account. The number –15 can be written as the product of the following numbers.

15=115$-15=-1 \cdot 15$ and 1+15=14$-1+15=14$ Notice that these are two different choices.

And also,

15=1(15)$-15=1 \cdot (-15)$ and 1+(15)=14$1+(-15)=-14$ Notice that these are two different choices.

15=(3)×5$-15=(-3) \times 5$ and (3)+5=2$(-3)+5=2$ This is the correct choice.

15=3×(5)$-15=3 \times (-5)$ and 3+(5)=2$3+(-5)=-2$

The answer is (x3)(x+5)$(x-3)(x+5)$.

Example 4: Factor x2+x+6$-x^2+x+6$.

Solution: First factor the common factor of –1 from each term in the trinomial. Factoring –1 changes the signs of each term in the expression.

x2+x+6=(x2x6)

We are looking for an answer that is a product of two parentheses (x± )(x± )$(x \pm \underline{\;\;\;\;\;\;\;\;\;\;\;} \ )(x \pm \underline{\;\;\;\;\;\;\;\;\;\;\;} \ )$.

Now our job is to factor x2x6$x^2-x-6$.

The number –6 can be written as the product of the following numbers.

&-6=2 \times (-3) \qquad and \qquad 2+(-3)=-1 \qquad This \ is \ the \ correct \ choice.

The answer is (x3)(x+2)$-(x-3)(x+2)$.

To Summarize:

A quadratic of the form x2+bx+c$x^2+bx+c$ factors as a product of two parenthesis (x+m)(x+n)$(x+m)(x+n)$.

• If b$b$ and c$c$ are positive then both m$m$ and n$n$ are positive.
• Example x2+8x+12$x^2+8x+12$ factors as (x+6)(x+2)$(x+6)(x+2)$.
• If b$b$ is negative and c$c$ is positive then both m$m$ and n$n$ are negative.
• Example x26x+8$x^2-6x+8$ factors as (x2)(x4)$(x-2)(x-4)$.
• If c$c$ is negative then either m$m$ is positive and n$n$ is negative or vice-versa.
• Example x2+2x15$x^2+2x-15$ factors as (x+5)(x3)$(x+5)(x-3)$.
• Example x2+34x35$x^2+34x-35$ factors as (x+35)(x1)$(x+35)(x-1)$.
• If a=1$a=-1$, factor a common factor of –1 from each term in the trinomial and then factor as usual. The answer will have the form (x+m)(x+n)$-(x+m)(x+n)$.
• Example x2+x+6$-x^2+x+6$ factors as (x3)(x+2)$-(x-3)(x+2)$.

## Practice Set

Sample explanations for some of the practice exercises below are available by viewing the following video. Note that there is not always a match between the number of the practice exercise in the video and the number of the practice exercise listed in the following exercise set.  However, the practice exercise is the same in both.

1. x2+10x+9$x^2+10x+9$
2. x2+15x+50$x^2+15x+50$
3. x2+10x+21$x^2+10x+21$
4. x2+16x+48$x^2+16x+48$
5. x211x+24$x^2-11x+24$
6. x213x+42$x^2-13x+42$
7. x214x+33$x^2-14x+33$
8. x29x+20$x^2-9x+20$
9. x2+5x14$x^2+5x-14$
10. x2+6x27$x^2+6x-27$
11. x2+7x78$x^2+7x-78$
12. x2+4x32$x^2+4x-32$
13. x212x45$x^2-12x-45$
14. x25x50$x^2-5x-50$
15. x23x40$x^2-3x-40$
16. x2x56$x^2-x-56$
17. x22x1$-x^2-2x-1$
18. x25x+24$-x^2-5x+24$
19. x2+18x72$-x^2+18x-72$
20. x2+25x150$-x^2+25x-150$
21. x2+21x+108$x^2+21x+108$
22. x2+11x30$-x^2+11x-30$
23. x2+12x64$x^2+12x-64$
24. x217x60$x^2-17x-60$

Mixed Review

1. Evaluate f(2)$f(2)$ when f(x)=12x26x+4$f(x)=\frac{1}{2} x^2-6x+4$.
2. The Nebraska Department of Roads collected the following data regarding mobile phone distractions in traffic crashes by teen drivers.
1. Plot the data as a scatter plot.
2. Fit a line to this data.
3. Predict the number of teenage traffic accidents attributable to cell phones in the year 2012.
Year (y$y$) Total (n$n$)
2002 41
2003 43
2004 47
2005 38
2006 36
2007 40
2008 42
2009 42
1. Simplify 405$\sqrt{405}$.
2. Graph the following on a number line: π,2,53,310,16$-\pi, \sqrt{2}, \frac{5}{3}, - \frac{3}{10}, \sqrt{16}$.
3. What is the multiplicative inverse of 94$\frac{9}{4}$?

## Quick Quiz

1. Name the following polynomial. State its degree and leading coefficient 6x2y4z+6x62y5+11xyz4$6x^2 y^4 z+6x^6-2y^5+11xyz^4$.
2. Simplify (a2b2c+11abc5)+(4abc53a2b2c+9abc)$(a^2 b^2 c+11abc^5 )+(4abc^5-3a^2 b^2 c+9abc)$.
3. A rectangular solid has dimensions (a+2)$(a+2)$ by (a+4)$(a+4)$ by (3a)$(3a)$. Find its volume.
4. Simplify 3hjk3(h2j4k+6hk2)$-3hjk^3 (h^2 j^4 k+6hk^2)$.
5. Find the solutions to (x3)(x+4)(2x1)=0$(x-3)(x+4)(2x-1)=0$.
6. Multiply (a9b)(a+9b)$(a-9b)(a+9b)$.

8 , 9

Feb 22, 2012

Dec 11, 2014