Wize High School Grade 11 Math Textbook > Factoring Polynomials

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).



PAGE BREAK
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

0:00 / 0:00

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   ✓​

PAGE BREAK
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   ✓​

PAGE BREAK
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

Mark Yourself Question

Grab a piece of paper and try this problem yourself.
When you're done, check the "I have answered this question" box below.
View the solution and report whether you got it right or wrong.

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.

I have answered this question

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

Factoring

Practice: Factoring
Factor x2−10x+21x^2-10x+21x2−10x+21

Factoring

Practice: Factoring
Factor x2−9x−10x^2-9x-10x2−9x−10

Wize High School Grade 11 Math Textbook > Factoring Polynomials

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

Popular Courses

Factoring Simple Trinomials $\bco{y=x^2+bx+c}$

Recall - Expanding Binomials

How do we Factor Simple Trinomials $\bco{y=x^2+bx+c}$ ?

Practice: Simple Trinomials

Example: Factoring Simple Trinomials

Practice: Simple Trinomials

Practice: Factoring Simple Trinomials

Practice: Factoring Simple Trinomials

Practice: Factoring Simple Trinomial

Extra Practice

Factoring

Factoring

Wize High School Grade 11 Math Textbook > Factoring Polynomials

Factoring Simple Trinomials y=x2+bx+cy=x^2+bx+cy=x2+bx+c

Popular Courses

Factoring Simple Trinomials y=x2+bx+c\bco{y=x^2+bx+c}y=x2+bx+c

Recall - Expanding Binomials

How do we Factor Simple Trinomials y=x2+bx+c\bco{y=x^2+bx+c}y=x2+bx+c?

Practice: Simple Trinomials

Example: Factoring Simple Trinomials

Practice: Simple Trinomials

Practice: Factoring Simple Trinomials

Practice: Factoring Simple Trinomials

Practice: Factoring Simple Trinomial

Extra Practice

Factoring

Factoring

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

Factoring Simple Trinomials $\bco{y=x^2+bx+c}$

How do we Factor Simple Trinomials $\bco{y=x^2+bx+c}$ ?