Factorising Trinomials Using the “ac” Method

Another way to factor trinomials of the form \(a{x}^{2}+bx+c\) is the “ac” method. (The “ac” method is sometimes called the grouping method.) The “ac” method is actually an extension of the methods you used in the last section to factor trinomials with leading coefficient one. This method is very structured (that is step-by-step), and it always works!

Example: How to Factor Trinomials Using the “ac” Method

Factor: \(6{x}^{2}+7x+2\).

Solution

This table lists the steps for factoring 6 x ^ 2 + 7 x + 2. The first step is to factor the GCF. This polynomial has none.

The second row states to find the product a c. Then, it lists a c as 6 times 2 = 12.

The third step is to find two numbers m and n in which m times n = a c and m + n = b. The middle column reads, “find two numbers that add to 7. Both factors must be positive”. The numbers are 3 and 4. 3 times 4 is 12 and 3 + 4 is 7.

The next step is to split the middle term using m and n. That is, to write 7 x as 3 x + 4 x. Therefore, 6 x ^ 2 + 7 x + 2 is rewritten as 6 x ^ 2 +3 x + 4 x + 2.

The next step is to factor by grouping. 3 x(2 x + 1) + 2(2 x + 1) then factor again (2 x + 1)(3 x + 2).

The last step is to check by multiplying. Multiply the factors (2 x + 1)(3 x + 2) to get 6 x ^ 2 + 7 x + 2.

Factor trinomials of the form using the “ac” method.

Factor any GCF.
Find the product ac.
Find two numbers m and n that:
\(\begin{array}{cccc}\text{Multiply to}\phantom{\rule{0.2em}{0ex}}ac\hfill & & & m·n=a·c\hfill \\ \text{Add to}\phantom{\rule{0.2em}{0ex}}b\hfill & & & m+n=b\hfill \end{array}\)
Split the middle term using m and n:
Factor by grouping.
Check by multiplying the factors.

When the third term of the trinomial is negative, the factors of the third term will have opposite signs.

Example

Factor: \(8{u}^{2}-17u-21\).

Solution

Is there a greatest common factor? No.
Find \(a\cdot c\).	\(a\cdot c\)
	\(8(-21)\)
	\(-168\)

Find two numbers that multiply to \(-168\) and add to \(-17\). The larger factor must be negative.

Factors of \(-168\)	Sum of factors
\(1,-168\)	\(1+(-168)=-167\)
\(2,-84\)	\(2+(-84)=-82\)
\(3,-56\)	\(3+(-56)=-53\)
\(4,-42\)	\(4+(-42)=-38\)
\(6,-28\)	\(6+(-28)=-22\)
\(7,-24\)	\(7+(-24)=-17\text{*}\)
\(8,-21\)	\(8+(-21)=-13\)

\(\begin{array}{cccc}\text{Split the middle term using}\phantom{\rule{0.2em}{0ex}}7u\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}-24u.\hfill & & & \hfill \phantom{\rule{2em}{0ex}}8{u}^{2}-\underset{\swarrow}{17}\underset{\searrow}{u}-21\hfill \\ & & & \hfill \phantom{\rule{2em}{0ex}}\underset{└\_\_\_\_\_\_\_\_┘}{8{u}^{2}+7u}\phantom{\rule{0.2em}{0ex}}\underset{└\_\_\_\_\_\_\_\_\_\_\_┘}{-24u-21}\hfill \\ \text{Factor by grouping.}\hfill & & & \hfill \phantom{\rule{2em}{0ex}}u(8u+7)-3(8u+7)\hfill \\ & & & \hfill \phantom{\rule{2em}{0ex}}(8u+7)(u-3)\hfill \\ \text{Check by multiplying.}\hfill & & & \\ \phantom{\rule{2.5em}{0ex}}(8u+7)(u-3)\hfill & & & \\ \phantom{\rule{2.5em}{0ex}}8{u}^{2}-24u+7u-21\hfill & & & \\ \phantom{\rule{2.5em}{0ex}}8{u}^{2}-17u-21\phantom{\rule{0.2em}{0ex}}✓\hfill & & & \end{array}\)

Example

Factor: \(2{x}^{2}+6x+5\).

Solution

Is there a greatest common factor? No.
Find \(a\cdot c\).	\(a\cdot c\)
	\(2(5)\)
	\(10\)

Find two numbers that multiply to 10 and add to 6.

Factors of \(10\)	Sum of factors
\(1,10\)	\(1+10=11\)
2, 5	\(2+5=7\)

There are no factors that multiply to 10 and add to 6. The polynomial is prime.

Don’t forget to look for a common factor!

Example

Factor: \(10{y}^{2}-55y+70\).

Solution

Is there a greatest common factor? Yes. The GCF is 5.
Factor it. Be careful to keep the factor of 5 all the way through the solution!
The trinomial inside the parentheses has a leading coefficient that is not 1.
Factor the trinomial.
Check by mulitplying all three factors.
\(5(2{y}^{2}-2y-4y+14)\)
\(5(2{y}^{2}-11y+14)\)
\(10{y}^{2}-55y+70✓\)

We can now update the Preliminary Factoring Strategy, as shown in the figure below and detailed in section titled "Choose a strategy to factor polynomials completely (updated)" below, to include trinomials of the form \(a{x}^{2}+bx+c\). Remember, some polynomials are prime and so they cannot be factored.

This figure has the strategy for factoring polynomials. At the top of the figure is GCF. Below this, there are three options. The first is binomial. The second is trinomial. Under trinomial there are x squared + b x + c and a x squared + b x +c. The two methods here are trial and error and the “a c” method. The third option is for more than three terms. It is grouping.

Choose a strategy to factor polynomials completely (updated).

Is there a greatest common factor?
- Factor it.
Is the polynomial a binomial, trinomial, or are there more than three terms?
- If it is a binomial, right now we have no method to factor it.
- If it is a trinomial of the form \({x}^{2}+bx+c\)
  Undo FOIL \((x\phantom{\rule{1em}{0ex}})(x\phantom{\rule{1em}{0ex}})\).
- If it is a trinomial of the form \(a{x}^{2}+bx+c\)
  Use Trial and Error or the “ac” method.
- If it has more than three terms
  Use the grouping method.
Check by multiplying the factors.

External Resource:

Access this website for additional instruction and practice with factoring trinomials of the form \(a{x}^{2}+bx+c\).

Factorising Trinomials Using the “ac” Method