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Wize High School Grade 10 Math Textbook > Factoring Polynomials

Factoring Harder Trinomials y=ax2+bx+cy=ax^2+bx+c

Factoring Simple Trinomials y=x2+bx+cy=x^2+bx+cPrevious Section
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Factoring Harder Trinomials y=ax2+bx+c\bco{y=ax^2+bx+c}

How do we Factor Harder Trinomials y=ax2+bx+c\bco{y=ax^2+bx+c}?

If you think about the FOIL or distributive method for multiplying binomials (px+m)(qx+n)(px+m)(qx+n), we see that:
  • (px)(qx)=ax2   →   the only x2 term(px)(qx)=ax^2~~~\to~~~\bcfi{\text{the only }x^2~\text{term}}
  • (px)(n)=pnx   →   part of the x term(px)(n)=pnx~~~\to~~~\bcfi{\text{part of the }x~\text{term}}
  • (m)(qx)=mqx   →   part of the x term(m)(qx)=mqx~~~\to~~~\bcfi{\text{part of the }x~\text{term}}
  • (m)(n)=mn   →   the only constant term(m)(n)=mn~~~\to~~~\bcfi{\text{the only constant term}}

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



Wize Tip
In summary, we are looking for four numbers p, q, m, np, ~q, ~m, ~n such that
  • p×q=a\bm{p \times q = a}
  • If aa is positive, then p, qp,\ q are both positive
  • If aa is negative, then one of p, qp,\ q is positive and the other is negative
  • m×n=c\bm{m\times n=c}
  • If cc is negative, then one of m, nm,~n is negative and the other is positive
  • If cc is positive and bb is positive, then both m, nm,~n are positive
  • If cc is positive and bb is negative, then both m, nm,~n are negative
  • the "crisscross" product betweeen p,qp, q and m,nm, n is bb -- this requires some trial and error!

Then, the factored form will be ax2+bx+c=(px+m)(qx+n)\large \boxed{ax^2+bx+c=(px+m)(qx+n)}.

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Example: Factoring Harder Trinomials

Factor fully 2x2−x−32x^2-x-3.

We need two numbers p, qp,\ q to multiply to a=2a=2 (since this is positive, both p, qp,\ q are positive):
  • 2×1=22\times1=2

We need two numbers m, nm,\ n to multiply to c=−3c=-3 (since this is negative, one of these numbers is positive, the other is negative):
  • −3×1=−3-3\times1=-3
  • 3×−1=−33\times-1=-3

Now we need to use some trial and error to figure out the order of these numbers:


OR


Combination 4 gives us the correct result 2x2−3x+2x−3=2x2−x−32x^2-3x+2x-3=2x^2-x-3.

So, the factored form is 2x2−x−3=(2x−3)(x+1)\boxed{2x^2-x-3=(2x-3)(x+1)}.

Practice: Factoring Harder Trinomials

Fill in the blanks to factor the following harder trinomials.

a) 4x2+5x−6=(4x+4x^2+5x-6=(4x+
)(x+)(x+
))

b) 12x2−23x+10=(3x+12x^2-23x+10=(3x+
)(4x+)(4x+
))

Practice: Factoring Harder Trinomials

Factor the quadratic expression 3x2−7x+23x^2-7x+2.
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Factoring Harder Trinomials: Decompose Strategy

We can use the following steps to decompose then factor a harder trinomial in the form y=ax2+bx+cy=ax^2+bx+c:
  1. Find two numbers m, nm,~n so that
  2. m×n=a×cm\times n=a\times c
  3. m+n=bm+n=b
  4. Rewrite bxbx as mx+nxmx+nx to decompose the trinomial into ax2+mx+nx+cax^2+mx+nx+c
  5. Use group common factoring to factor the polynomial ax2+mx+nx+cax^2+mx+nx+c


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Example
Factor 2x2−x−32x^2-x-3.

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

1. Find two numbers m,nm, n so that:
  • m×n=2×(−3)=−6m\times n=2\times (-3)=-6
  • m+n=−1m+n=-1
These two numbers are −3-3 and 22

2. Decompose the polynomial:
2x2−x−3=2x2−3x+2x−3\begin{array}{cccc} &2x^2&\colorTwo{-x}&-3\\[1em] =&2x^2&\colorTwo{-3x+2x}&-3 \end{array}

3. Use group common factoring to factor this new polynomial:
2x2−3x⏟Group 1+2x−3⏟Group 2=x(2x−3)+1(2x−3)=x(2x−3)+1(2x−3)\begin{array}{cccc} &\underbrace{2x^2-3x}_\text{Group 1}&+&\underbrace{2x-3}_\text{Group 2}\\[2em] =&x(2x-3)&+&1(2x-3)\\[1em] =&x\bcfi{(2x-3)}&+&1\bcfi{(2x-3)}\\[1em] \end{array}
=(2x−3)(x+1)\begin{array}{cc} =&\bcfi{(2x-3)}(x+1) \end{array}

So, the factored form is 2x2−x−3=(2x−3)(x+1)\boxed{2x^2-x-3=(2x-3)(x+1)}.

*This method doesn't involve the crisscross trial and error.

Practice: Factoring Harder Trinomials

Factor the polynomial 6x2+5x−66x^2+5x-6.

Practice: Factoring Harder Trinomials

Factor the following polynomials fully.

a) 36x2+39x−1236x^2+39x-12

b) 8pt2−26pt+15p8pt^2-26pt+15p

Practice: Factoring Harder Trinomials

Factor the following polynomials.

a) 5x2+22x+85x^2+22x+8

b) 6y2+13y−156y^2+13y-15

c) 12h2−34h+1012h^2-34h+10

d) −4p2−8p+5-4p^2-8p+5

e) 6dt2+21dt−12d6dt^2+21dt-12d

f) 4x2−94x^2-9