Wize High School Algebra I Textbook (Common Core) > Factoring Polynomials

Factoring Simple Trinomials $y=x^2+bx+c$

y=ax^2+bx+c

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Factoring Simple Trinomials y=x2+bx+c\bco{y=x^2+bx+c}y=x2+bx+c
Recall - Expanding Binomials
When we expand and multiply two binomials together, we sometimes get a trinomial:

(x+1)(x+2)= x2+2x+x+2= x2+3x+2\large\begin{aligned}
&(x+1)(x+2)\\
=&~x^2+2x+x+2\\
=&~x^2+3x+2
\end{aligned}==​(x+1)(x+2) x2+2x+x+2 x2+3x+2​

So, we should be able to work backwards and factor a trinomial of the form y=x2+bx+cy=x^2+bx+cy=x2+bx+c into two linear factors y=(x+m)(x+n)y=(x+m)(x+n)y=(x+m)(x+n).



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How do we Factor Simple Trinomials y=x2+bx+c\bco{y=x^2+bx+c}y=x2+bx+c?
If you think about the FOIL or distributive method for multiplying binomials (x+m)(x+n)(x+m)(x+n)(x+m)(x+n), we see that:
(x)(x)=x2   →   the only x2 term(x)(x)=x^2~~~\to~~~\bcfi{\text{the only }x^2~\text{term}}(x)(x)=x2   →   the only x2 term 
(x)(n)=nx   →   part of the x term(x)(n)=nx~~~\to~~~\bcfi{\text{part of the }x~\text{term}}(x)(n)=nx   →   part of the x term
(m)(x)=mx   →   part of the x term(m)(x)=mx~~~\to~~~\bcfi{\text{part of the }x~\text{term}}(m)(x)=mx   →   part of the x term
(m)(n)=mn   →   the only constant term(m)(n)=mn~~~\to~~~\bcfi{\text{the only constant term}}(m)(n)=mn   →   the only constant term

We can use this factoring box tool to help us visualize how to factor a simple trinomial:



Wize Tip
In summary, we are looking for two numbers mmm and nnn such that
m×n=c\bm{m \times n =c}m×n=c
m+n=b\bm{m+n=b}m+n=b
The order of mmm and nnn doesn't matter!

Then, the factored form will be x2+bx+c=(x+m)(x+n)\large \boxed{x^2+bx+c=(x+m)(x+n)}x2+bx+c=(x+m)(x+n)​.

Practice: Simple Trinomials
For each of the following trinomials in the form y=ax2+bx+cy=ax^2+bx+cy=ax2+bx+c,

i) identify the values of aaa, bbb, and ccc

ii) write down all integer pairs mmm and nnn that multiply to the value of ccc 
(in other words, write down all possible integers mmm and nnn so that m×n=cm\times n=cm×n=c)

y=x2+6x+5y=x^2+6x+5y=x2+6x+5

a=

b=

c=

ii) How many unique integer pairs

m

and

n

are there? (for example

m=1

and

n=2

is the same as

m=2

and

n=1

)

I don't know

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Example: Factoring Simple Trinomials
Factor the following polynomials.

a) y=x2+3x+2y=x^2+3x+2y=x2+3x+2

First list out: a=1, b=3, c=2a=1,~b=3, ~c=2a=1, b=3, c=2.

We need
m×n=c   →   m×n=2m\times n=c~~~\to~~~m\times n=2m×n=c   →   m×n=2
m+n=b   →   m+n=3m+n=b~~~\to~~~m+n=3m+n=b   →   m+n=3
Our two numbers are m=1m=1m=1 and n=2n=2n=2 (the order doesn't matter!)

So, y=x2+3x+2=(x+1)(x+2)\begin{aligned}
y&=x^2+3x+2\\
&=\boxed{(x+1)(x+2)}
\end{aligned}y​=x2+3x+2=(x+1)(x+2)​​

Check: (x+1)(x+2)=(x)(x)+(x)(2)+(1)(x)+(1)(2)=x2+2x+x+2=x2+3x+2   ✓\begin{aligned}
&(x+1)(x+2)\\
=&(x)(x)+(x)(2)+(1)(x)+(1)(2)\\
=&x^2+2x+x+2\\
=&x^2+3x+2~~~\checkmark
\end{aligned}===​(x+1)(x+2)(x)(x)+(x)(2)+(1)(x)+(1)(2)x2+2x+x+2x2+3x+2   ✓​

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b) y=x2−2x−15y=x^2-2x-15y=x2−2x−15

First list out: a=1, b=−2, c=−15a=1,~b=-2, ~c=-15a=1, b=−2, c=−15.

We need
m×n=c   →   m×n=−15m\times n=c~~~\to~~~m\times n=-15m×n=c   →   m×n=−15
m+n=b   →   m+n=−2m+n=b~~~\to~~~m+n=-2m+n=b   →   m+n=−2
Our two numbers are m=−5m=-5m=−5 and n=3n=3n=3 (the order doesn't matter!)

So, y=x2−2x−15=(x−5)(x+3)\begin{aligned}
y&=x^2-2x-15\\
&=\boxed{(x-5)(x+3)}
\end{aligned}y​=x2−2x−15=(x−5)(x+3)​​

Check: (x−5)(x+3)=(x)(x)+(x)(3)+(−5)(x)+(−5)(3)=x2+3x−5x−15=x2−2x−15   ✓\begin{aligned}
&(x-5)(x+3)\\
=&(x)(x)+(x)(3)+(-5)(x)+(-5)(3)\\
=&x^2+3x-5x-15\\
=&x^2-2x-15~~~\checkmark
\end{aligned}===​(x−5)(x+3)(x)(x)+(x)(3)+(−5)(x)+(−5)(3)x2+3x−5x−15x2−2x−15   ✓​

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c) y=x2−8x+12y=x^2-8x+12y=x2−8x+12

First list out: a=1, b=−8, c=12a=1,~b=-8, ~c=12a=1, b=−8, c=12.

We need
m×n=c   →   m×n=12m\times n=c~~~\to~~~m\times n=12m×n=c   →   m×n=12
m+n=b   →   m+n=−8m+n=b~~~\to~~~m+n=-8m+n=b   →   m+n=−8
Our two numbers are m=−6m=-6m=−6 and n=−2n=-2n=−2 (the order doesn't matter!)

So, y=x2−8x+12=(x−6)(x−2)\begin{aligned}
y&=x^2-8x+12\\
&=\boxed{(x-6)(x-2)}
\end{aligned}y​=x2−8x+12=(x−6)(x−2)​​

Check: (x−6)(x−2)=(x)(x)+(x)(−2)+(−6)(x)+(−6)(−2)=x2−2x−6x+12=x2−8x+12   ✓\begin{aligned}
&(x-6)(x-2)\\
=&(x)(x)+(x)(-2)+(-6)(x)+(-6)(-2)\\
=&x^2-2x-6x+12\\
=&x^2-8x+12~~~\checkmark
\end{aligned}===​(x−6)(x−2)(x)(x)+(x)(−2)+(−6)(x)+(−6)(−2)x2−2x−6x+12x2−8x+12   ✓​

Practice: Simple Trinomials
Factor each of the following trinomials.

a) y=x2+6x+5y=x^2+6x+5y=x2+6x+5

b) y=x2+5x−14y=x^2+5x-14y=x2+5x−14

c) y=x2−11x+24y=x^2-11x+24y=x2−11x+24

a) Enter the factored trinomial in the form

y=(x...)(x...)

b) Enter the factored trinomial in the form

y=(x...)(x...)

c) Enter the factored trinomial in the form

y=(x...)(x...)

I don't know

Practice: Factoring Simple Trinomials
Factor the following polynomials.

a) y=2x2+10x−100y=2x^2+10x-100y=2x2+10x−100

b) y=x2−4y=x^2-4y=x2−4

a) Enter the factored polynomial in the form

y=...

b) Enter the factored polynomial in the form

y=(...)(...)

I don't know

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Practice: Factoring Simple Trinomials
Find the roots (zeros) of the quadratic equation y=x2−10x+16y=x^2-10x+16y=x2−10x+16. Then sketch the graph.

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Practice: Factoring Simple Trinomial
Factor the following polynomials.

a) x2+4x+3x^2+4x+3x2+4x+3

b) t2+7t+12t^2+7t+12t2+7t+12

c) n2−5n−6n^2-5n-6n2−5n−6

d) y2+6y−40y^2+6y-40y2+6y−40

e) z2+5z−14z^2+5z-14z2+5z−14

f) m2−10m−24m^2-10m-24m2−10m−24

I don't know

Extra Practice

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Wize High School Algebra I Textbook (Common Core) > Factoring Polynomials

Factoring Simple Trinomials $y=x^2+bx+c$

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