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Factor Trinomials of the Form ax

Recognize a preliminary strategy to factor polynomials completely

Recognize a Preliminary Strategy for Factoring

Let’s summarize where we are so far with factoring polynomials. In the first two sections of this chapter, we used three methods of factoring: factoring the GCF, factoring by grouping, and factoring a trinomial by “undoing” FOIL. More methods will follow as you continue in this chapter, as well as later in your studies of algebra.

How will you know when to use each factoring method? As you learn more methods of factoring, how will you know when to apply each method and not get them confused? It will help to organize the factoring methods into a strategy that can guide you to use the correct method.

As you start to factor a polynomial, always ask first, “Is there a greatest common factor?” If there is, factor it first.

The next thing to consider is the type of polynomial. How many terms does it have? Is it a binomial? A trinomial? Or does it have more than three terms?

If it is a trinomial where the leading coefficient is one, \({x}^{2}+bx+c\), use the “undo FOIL” method.

If it has more than three terms, try the grouping method. This is the only method to use for polynomials of more than three terms.

Some polynomials cannot be factored. They are called “prime.”

Example

Try it.

Identify the best method to use to factor each polynomial.

ⓐ \(6{y}^{2}-72\)
ⓑ \({r}^{2}-10r-24\)
ⓒ \({p}^{2}+5p+pq+5q\)

Solution

ⓐ

	\(6{y}^{2}-72\)
Is there a greatest common factor?	Yes, 6.
Factor out the 6.	\(6({y}^{2}-12)\)
Is it a binomial, trinomial, or are there more than 3 terms?	Binomial, we have no method to factor binomials yet.

ⓑ

	\({r}^{2}-10r-24\)
Is there a greatest common factor?	No, there is no common factor.
Is it a binomial, trinomial, or are there more than three terms?	Trinomial, with leading coefficient 1, so “undo” FOIL.

ⓒ

	\({p}^{2}+5p+pq+5q\)
Is there a greatest common factor?	No, there is no common factor.
Is it a binomial, trinomial, or are there more than three terms?	More than three terms, so factor using grouping.

Condensed — the full section is in OpenStax Elementary Algebra 2e.

Factor Trinomials of the form

Now that we have organized what we’ve covered so far, we are ready to factor trinomials whose leading coefficient is not 1, trinomials of the form \(a{x}^{2}+bx+c\).

Remember to always check for a GCF first! Sometimes, after you factor the GCF, the leading coefficient of the trinomial becomes 1 and you can factor it by the methods in the last section. Let’s do a few examples to see how this works.

Watch out for the signs in the next two examples.

Example

Try it.

Factor completely: \(2{n}^{2}-8n-42\).

Solution

Use the preliminary strategy.

Is there a greatest common factor?	\(2{n}^{2}-8n-42\)
Yes, GCF = 2. Factor it out.	\(2({n}^{2}-4n-21)\)

Inside the parentheses, is it a binomial, trinomial, or are there more than three terms?

It is a trinomial whose coefficient is 1, so undo FOIL.	\(2(n\ )(n\ )\)
Use 3 and −7 as the last terms of the binomials.	\(2(n+3)(n-7)\)

Factors of \(-21\)	Sum of factors
\(1,-21\)	\(1+(-21)=-20\)
\(3,-7\)	\(3+(-7)=-4\text{*}\)

Check.

\(\ 2(n+3)(n-7)\)

\(\ 2({n}^{2}-7n+3n-21)\)

\(\ 2({n}^{2}-4n-21)\)

\(\ 2{n}^{2}-8n-42\ ✓\)

Example

Try it.

Factor completely: \(4{y}^{2}-36y+56\).

Solution

Use the preliminary strategy.

Is there a greatest common factor?	\(4{y}^{2}-36y+56\)
Yes, GCF = 4. Factor it.	\(4({y}^{2}-9y+14)\)
Inside the parentheses, is it a binomial, trinomial, or are there more than three terms?
It is a trinomial whose coefficient is 1. So undo FOIL.	\(4(y\ )(y\ )\)
Use a table like the one below to find two numbers that multiply to 14 and add to −9.
Both factors of 14 must be negative.	\(4(y-2)(y-7)\)

Factors of \(14\)	Sum of factors
\(-1,-14\)	\(-1+(-14)=-15\)
\(-2,-7\)	\(-2+(-7)=-9\text{*}\)

Check.

\(\ 4(y-2)(y-7)\)

\(\ 4({y}^{2}-7y-2y+14)\)

\(\ 4({y}^{2}-9y+14)\)

\(\ 4{y}^{2}-36y+56\ ✓\)

In the next example the GCF will include a variable.

Example

Try it.

Factor completely: \(4{u}^{3}+16{u}^{2}-20u\).

Solution

Use the preliminary strategy.

Is there a greatest common factor?	\(4{u}^{3}+16{u}^{2}-20u\)
Yes, GCF = 4u. Factor it.	\(4u({u}^{2}+4u-5)\)
Binomial, trinomial, or more than three terms?
It is a trinomial. So “undo FOIL.”	\(4u(u\ )(u\ )\)
Use a table like the table below to find two numbers that multiply to −5 and add to 4.	\(4u(u-1)(u+5)\)

Factors of \(-5\)	Sum of factors
\(-1,5\)	\(-1+5=4\text{*}\)
\(1,-5\)	\(1+(-5)=-4\)

Check.

\(\ 4u(u-1)(u+5)\)

\(\ 4u({u}^{2}+5u-u-5)\)

\(\ 4u({u}^{2}+4u-5)\)

\(\ 4{u}^{3}+16{u}^{2}-20u\ ✓\)

Factor Trinomials using Trial and Error

What happens when the leading coefficient is not 1 and there is no GCF? There are several methods that can be used to factor these trinomials. First we will use the Trial and Error method.

Let’s factor the trinomial \(3{x}^{2}+5x+2\).

From our earlier work we expect this will factor into two binomials.

\[\begin{array}{l}3{x}^{2}+5x+2 \\ (\ )(\ )\end{array}\]

We know the first terms of the binomial factors will multiply to give us \(3{x}^{2}\). The only factors of \(3{x}^{2}\) are \(1x,3x\). We can place them in the binomials.

Check. Does \(1x\cdot 3x=3{x}^{2}\)?

We know the last terms of the binomials will multiply to 2. Since this trinomial has all positive terms, we only need to consider positive factors. The only factors of 2 are 1 and 2. But we now have two cases to consider as it will make a difference if we write 1, 2, or 2, 1.

Which factors are correct? To decide that, we multiply the inner and outer terms.

\[\begin{array}{l}(x+1)(3x+2) \\ 3{x}^{2}+2x+3x+2 \\ 3{x}^{2}+5x+2\ ✓\end{array}\]\[\begin{array}{l} \\ 3{x}^{2}+5x+2 \\ (x+1)(3x+2)\end{array}\]

How to Factor Trinomials of the Form

Try it.

Factor completely: \(3{y}^{2}+22y+7\).

Solution

Example

Try it.

Factor completely: \(6{b}^{2}-13b+5\).

Solution

The trinomial is already in descending order.
Find the factors of the first term.
Find the factors of the last term. Consider the signs. Since the last term, 5 is positive its factors must both be positive or both be negative. The coefficient of the middle term is negative, so we use the negative factors.

Consider all the combinations of factors.

\(6{b}^{2}-13b+5\)
Possible factors	Product
\((b-1)(6b-5)\)	\(6{b}^{2}-11b+5\)
\((b-5)(6b-1)\)	\(6{b}^{2}-31b+5\)
\((2b-1)(3b-5)\)	\(6{b}^{2}-13b+5\) *
\((2b-5)(3b-1)\)	\(6{b}^{2}-17b+5\)

The correct factors are those whose product is the original trinomial.	\((2b-1)(3b-5)\)
Check by multiplying. \(\begin{array}{l} \\ \\ (2b-1)(3b-5) \\ 6{b}^{2}-10b-3b+5 \\ 6{b}^{2}-13b+5\ ✓\end{array}\)

Condensed — the full section is in OpenStax Elementary Algebra 2e.

Factor 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!

How to Factor Trinomials Using the “ac” Method

Try it.

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

Solution

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

Example

Try it.

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

Solution

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

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

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

Split the middle term using 7u and −24u.	\(\begin{array}{l} \\ \\ \\ 8{u}^{2}-\underset{\text{↙}}{17}\underset{\text{↘}}{u}-21 \\ \underset{⎵}{8{u}^{2}+7u}\ \underset{⎵}{-24u-21}\end{array}\)
Factor by grouping.	\(\begin{array}{l} \\ u(8u+7)-3(8u+7) \\ (8u+7)(u-3)\end{array}\)
Check by multiplying. \(\begin{array}{l} \\ \\ (8u+7)(u-3) \\ 8{u}^{2}-24u+7u-21 \\ 8{u}^{2}-17u-21\ ✓\end{array}\)

Example

Try it.

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

Solution

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

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

Try it.

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.
\(5(2{y}^{2}-2y-4y+14)\)
\(5(2{y}^{2}-11y+14)\)
\(10{y}^{2}-55y+70✓\)

Condensed — the full section is in OpenStax Elementary Algebra 2e.

Key Concepts

Factor Trinomials of the Form \(a{x}^{2}+bx+c\) using Trial and Error: See .
1. Write the trinomial in descending order of degrees.
2. Find all the factor pairs of the first term.
3. Find all the factor pairs of the third term.
4. Test all the possible combinations of the factors until the correct product is found.
5. Check by multiplying.
Factor Trinomials of the Form \(a{x}^{2}+bx+c\) Using the “ac” Method: See .
1. Factor any GCF.
2. Find the product ac.
3. Find two numbers m and n that:
  \(\begin{array}{llll}\text{Multiply to}\ ac & & & m\cdot n=a\cdot c \\ \text{Add to}\ b & & & m+n=b\end{array}\)
4. Split the middle term using m and n:
5. Factor by grouping.
6. Check by multiplying the factors.
Choose a strategy to factor polynomials completely (updated):
1. Is there a greatest common factor? Factor it.
2. 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 \((x\ )(x\ )\).
  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.
3. Check by multiplying the factors.

Factor Trinomials of the Form ax

Recognize a Preliminary Strategy to Factor Polynomials Completely

In the following exercises, identify the best method to use to factor each polynomial.

Try it.

ⓐ \(10{q}^{2}+50\)
ⓑ \({a}^{2}-5a-14\)
ⓒ \(uv+2u+3v+6\)

Solution

ⓐ factor the GCF, binomial ⓑ Undo FOIL ⓒ factor by grouping

Try it.

ⓐ \({n}^{2}+10n+24\)
ⓑ \(8{u}^{2}+16\)
ⓒ \(pq+5p+2q+10\)

Try it.

ⓐ \({x}^{2}+4x-21\)
ⓑ \(ab+10b+4a+40\)
ⓒ \(6{c}^{2}+24\)

Solution

ⓐ undo FOIL ⓑ factor by grouping ⓒ factor the GCF, binomial

Try it.

ⓐ \(20{x}^{2}+100\)
ⓑ \(uv+6u+4v+24\)
ⓒ \({y}^{2}-8y+15\)

Factor Trinomials of the form \(a{x}^{2}+bx+c\) with a GCF

In the following exercises, factor completely.

Try it.

\(5{x}^{2}+35x+30\)

Solution

\(5(x+1)(x+6)\)

Try it.

\(12{s}^{2}+24s+12\)

Try it.

\(2{z}^{2}-2z-24\)

Solution

\(2(z-4)(z+3)\)

Try it.

\(3{u}^{2}-12u-36\)

Try it.

\(7{v}^{2}-63v+56\)

Solution

\(7(v-1)(v-8)\)

Try it.

\(5{w}^{2}-30w+45\)

Try it.

\({p}^{3}-8{p}^{2}-20p\)

Solution

\(p(p-10)(p+2)\)

Try it.

\({q}^{3}-5{q}^{2}-24q\)

Try it.

\(3{m}^{3}-21{m}^{2}+30m\)

Solution

\(3m(m-5)(m-2)\)

Try it.

\(11{n}^{3}-55{n}^{2}+44n\)

Try it.

\(5{x}^{4}+10{x}^{3}-75{x}^{2}\)

Solution

\(5{x}^{2}(x-3)(x+5)\)

Try it.

\(6{y}^{4}+12{y}^{3}-48{y}^{2}\)

Factor Trinomials Using Trial and Error

In the following exercises, factor.

Try it.

\(2{t}^{2}+7t+5\)

Solution

\((2t+5)(t+1)\)

Try it.

\(5{y}^{2}+16y+11\)

Try it.

\(11{x}^{2}+34x+3\)

Solution

\((11x+1)(x+3)\)

Try it.

\(7{b}^{2}+50b+7\)

Try it.

\(4{w}^{2}-5w+1\)

Solution

\((4w-1)(w-1)\)

Try it.

\(5{x}^{2}-17x+6\)

Try it.

\(6{p}^{2}-19p+10\)

Solution

\((3p-2)(2p-5)\)

Try it.

\(21{m}^{2}-29m+10\)

Try it.

\(4{q}^{2}-7q-2\)

Solution

\((4q+1)(q-2)\)

Try it.

\(10{y}^{2}-53y-11\)

Try it.

\(4{p}^{2}+17p-15\)

Solution

\((4p-3)(p+5)\)

Try it.

\(6{u}^{2}+5u-14\)

Try it.

\(16{x}^{2}-32x+16\)

Solution

\(16(x-1)(x-1)\)

Try it.

\(81{a}^{2}+153a-18\)

Try it.

\(30{q}^{3}+140{q}^{2}+80q\)

Solution

\(10q(3q+2)(q+4)\)

Try it.

\(5{y}^{3}+30{y}^{2}-35y\)

Factor Trinomials using the ‘ac’ Method

In the following exercises, factor.

Try it.

\(5{n}^{2}+21n+4\)

Solution

\((5n+1)(n+4)\)

Try it.

\(8{w}^{2}+25w+3\)

Try it.

\(9{z}^{2}+15z+4\)

Solution

\((3z+1)(3z+4)\)

Try it.

\(3{m}^{2}+26m+48\)

Try it.

\(4{k}^{2}-16k+15\)

Solution

\((2k-3)(2k-5)\)

Try it.

\(4{q}^{2}-9q+5\)

Try it.

\(5{s}^{2}-9s+4\)

Solution

\((5s-4)(s-1)\)

Try it.

\(4{r}^{2}-20r+25\)

Try it.

\(6{y}^{2}+y-15\)

Solution

\((3y+5)(2y-3)\)

Try it.

\(6{p}^{2}+p-22\)

Try it.

\(2{n}^{2}-27n-45\)

Solution

\((2n+3)(n-15)\)

Try it.

\(12{z}^{2}-41z-11\)

Try it.

\(3{x}^{2}+5x+4\)

Solution

prime

Try it.

\(4{y}^{2}+15y+6\)

Try it.

\(60{y}^{2}+290y-50\)

Solution

\(10(6y-1)(y+5)\)

Try it.

\(6{u}^{2}-46u-16\)

Try it.

\(48{z}^{3}-102{z}^{2}-45z\)

Solution

\(3z(8z+3)(2z-5)\)

Try it.

\(90{n}^{3}+42{n}^{2}-216n\)

Try it.

\(16{s}^{2}+40s+24\)

Solution

\(8(2s+3)(s+1)\)

Try it.

\(24{p}^{2}+160p+96\)

Try it.

\(48{y}^{2}+12y-36\)

Solution

\(12(4y-3)(y+1)\)

Try it.

\(30{x}^{2}+105x-60\)

Mixed Practice

In the following exercises, factor.

Try it.

\(12{y}^{2}-29y+14\)

Solution

\((4y-7)(3y-2)\)

Try it.

\(12{x}^{2}+36y-24z\)

Try it.

\({a}^{2}-a-20\)

Solution

\((a-5)(a+4)\)

Try it.

\({m}^{2}-m-12\)

Try it.

\(6{n}^{2}+5n-4\)

Solution

\((2n-1)(3n+4)\)

Try it.

\(12{y}^{2}-37y+21\)

Try it.

\(2{p}^{2}+4p+3\)

Solution

prime

Try it.

\(3{q}^{2}+6q+2\)

Try it.

\(13{z}^{2}+39z-26\)

Solution

\(13({z}^{2}+3z-2)\)

Try it.

\(5{r}^{2}+25r+30\)

Try it.

\({x}^{2}+3x-28\)

Solution

\((x+7)(x-4)\)

Try it.

\(6{u}^{2}+7u-5\)

Try it.

\(3{p}^{2}+21p\)

Solution

\(3p(p+7)\)

Try it.

\(7{x}^{2}-21x\)

Try it.

\(6{r}^{2}+30r+36\)

Solution

\(6(r+2)(r+3)\)

Try it.

\(18{m}^{2}+15m+3\)

Try it.

\(24{n}^{2}+20n+4\)

Solution

\(4(2n+1)(3n+1)\)

Try it.

\(4{a}^{2}+5a+2\)

Try it.

\({x}^{2}+2x-24\)

Solution

\((x+6)(x-4)\)

Try it.

\(2{b}^{2}-7b+4\)

Condensed — the full section is in OpenStax Elementary Algebra 2e.

Practice (40)

Try each one on paper first. Reveal the answer to check; verified ones can be opened in the solver for every step.

Find the GCF of \(45{p}^{2}\ \text{and}\ 30{p}^{6}\).
If you missed this problem, review .

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\(15{p}^{2}\)
Multiply \((3y+4)(2y+5)\).
If you missed this problem, review .

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\(6{y}^{2}+23y+20\)
Combine like terms \(12{x}^{2}+3x+5x+9\).
If you missed this problem, review .

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\(12{x}^{2}+8x+9\)

Identify the best method to use to factor each polynomial.

ⓐ \(6{y}^{2}-72\)
ⓑ \({r}^{2}-10r-24\)

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ⓐ

	\(6{y}^{2}-72\)
Is there a greatest common factor?	Yes, 6.
Factor out the 6.	\(6({y}^{2}-12)\)
Is it a binomial, trinomial, or are there more than 3 terms?	Binomial, we have no method to factor binomials yet.

ⓑ

	\({r}^{2}-10r-24\)
Is there a greatest common factor?	No, there is no common factor.
Is it a binomial, trinomial, or are there more than three terms?	Trinomial, with leading coefficient 1, so “undo” FOIL.

ⓒ

	\({p}^{2}+5p+pq+5q\)
Is there a greatest common factor?	No, there is no common factor.
Is it a binomial, trinomial, or are there more than three terms?	More than three terms, so factor using grouping.

Identify the best method to use to factor each polynomial:
1. ⓐ \(4{y}^{2}+32\)
2. ⓑ \({y}^{2}+10y+21\)
3. ⓒ \(yz+2y+3z+6\)
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ⓐ no method ⓑ undo using FOIL ⓒ factor with grouping
Identify the best method to use to factor each polynomial:
1. ⓐ \(ab+a+4b+4\)
2. ⓑ \(3{k}^{2}+15\)
3. ⓒ \({p}^{2}+9p+8\)
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ⓐ factor using grouping ⓑ no method ⓒ undo using FOIL
Factor completely: \(2{n}^{2}-8n-42\).

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Use the preliminary strategy.
Is there a greatest common factor? \(2{n}^{2}-8n-42\)
Yes, GCF = 2. Factor it out. \(2({n}^{2}-4n-21)\)

Inside the parentheses, is it a binomial, trinomial, or are there more than three terms?
It is a trinomial whose coefficient is 1, so undo FOIL. \(2(n\ )(n\ )\)
Use 3 and −7 as the last terms of the binomials. \(2(n+3)(n-7)\)

Factors of \(-21\) Sum of factors
\(1,-21\) \(1+(-21)=-20\)
\(3,-7\) \(3+(-7)=-4\text{*}\)
Check.
\(\ 2(n+3)(n-7)\)
\(\ 2({n}^{2}-7n+3n-21)\)
\(\ 2({n}^{2}-4n-21)\)
\(\ 2{n}^{2}-8n-42\ ✓\)
Factor completely: \(4{m}^{2}-4m-8\).

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\(4(m+1)(m-2)\)
Factor completely: \(5{k}^{2}-15k-50\).

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\(5(k+2)(k-5)\)

Factor completely: \(4{y}^{2}-36y+56\).

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Use the preliminary strategy.

Is there a greatest common factor?	\(4{y}^{2}-36y+56\)
Yes, GCF = 4. Factor it.	\(4({y}^{2}-9y+14)\)
Inside the parentheses, is it a binomial, trinomial, or are there more than three terms?
It is a trinomial whose coefficient is 1. So undo FOIL.	\(4(y\ )(y\ )\)
Use a table like the one below to find two numbers that multiply to 14 and add to −9.
Both factors of 14 must be negative.	\(4(y-2)(y-7)\)

Factors of \(14\)	Sum of factors
\(-1,-14\)	\(-1+(-14)=-15\)
\(-2,-7\)	\(-2+(-7)=-9\text{*}\)

Check.

\(\ 4(y-2)(y-7)\)

\(\ 4({y}^{2}-7y-2y+14)\)

\(\ 4({y}^{2}-9y+14)\)

\(\ 4{y}^{2}-36y+56\ ✓\)

Factor completely: \(3{r}^{2}-9r+6\).

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\(3(r-1)(r-2)\)
Factor completely: \(2{t}^{2}-10t+12\).

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\(2(t-2)(t-3)\)

Factor completely: \(4{u}^{3}+16{u}^{2}-20u\).

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Use the preliminary strategy.

Is there a greatest common factor?	\(4{u}^{3}+16{u}^{2}-20u\)
Yes, GCF = 4u. Factor it.	\(4u({u}^{2}+4u-5)\)
Binomial, trinomial, or more than three terms?
It is a trinomial. So “undo FOIL.”	\(4u(u\ )(u\ )\)
Use a table like the table below to find two numbers that multiply to −5 and add to 4.	\(4u(u-1)(u+5)\)

Factors of \(-5\)	Sum of factors
\(-1,5\)	\(-1+5=4\text{*}\)
\(1,-5\)	\(1+(-5)=-4\)

Check.

\(\ 4u(u-1)(u+5)\)

\(\ 4u({u}^{2}+5u-u-5)\)

\(\ 4u({u}^{2}+4u-5)\)

\(\ 4{u}^{3}+16{u}^{2}-20u\ ✓\)

Factor completely: \(5{x}^{3}+15{x}^{2}-20x\).

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\(5x(x-1)(x+4)\)
Factor completely: \(6{y}^{3}+18{y}^{2}-60y\).

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\(6y(y-2)(y+5)\)
Factor completely: \(3{y}^{2}+22y+7\).
Factor completely: \(2{a}^{2}+5a+3\).

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\((a+1)(2a+3)\)
Factor completely: \(4{b}^{2}+5b+1\).

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\((b+1)(4b+1)\)

Factor completely: \(6{b}^{2}-13b+5\).

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The trinomial is already in descending order.
Find the factors of the first term.
Find the factors of the last term. Consider the signs. Since the last term, 5 is positive its factors must both be positive or both be negative. The coefficient of the middle term is negative, so we use the negative factors.

Consider all the combinations of factors.

\(6{b}^{2}-13b+5\)
Possible factors	Product
\((b-1)(6b-5)\)	\(6{b}^{2}-11b+5\)
\((b-5)(6b-1)\)	\(6{b}^{2}-31b+5\)
\((2b-1)(3b-5)\)	\(6{b}^{2}-13b+5\) *
\((2b-5)(3b-1)\)	\(6{b}^{2}-17b+5\)

The correct factors are those whose product is the original trinomial.	\((2b-1)(3b-5)\)
Check by multiplying. \(\begin{array}{l} \\ \\ (2b-1)(3b-5) \\ 6{b}^{2}-10b-3b+5 \\ 6{b}^{2}-13b+5\ ✓\end{array}\)

Factor completely: \(8{x}^{2}-14x+3\).

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\((2x-3)(4x-1)\)
Factor completely: \(10{y}^{2}-37y+7\).

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\((2y-7)(5y-1)\)

Factor completely: \(14{x}^{2}-47x-7\).

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The trinomial is already in descending order.
Find the factors of the first term.
Find the factors of the last term. Consider the signs. Since it is negative, one factor must be positive and one negative.

Consider all the combinations of factors. We use each pair of the factors of \(14{x}^{2}\) with each pair of factors of \(-7.\)

Factors of \(14{x}^{2}\)	Pair with	Factors of \(-7\)
\(x\), \(14x\)		\(1\), \(-7\) \(-7\), \(1\) (reverse order)
\(x\), \(14x\)		\(-1\), \(7\) \(7\), \(-1\) (reverse order)
\(2x,7x\)		\(1\), \(-7\) \(-7\), \(1\) (reverse order)
\(2x,7x\)		\(-1\), \(7\) \(7\), \(-1\) (reverse order)

These pairings lead to the following eight combinations.

The correct factors are those whose product is the original trinomial.	\((2x-7)(7x+1)\)
Check by multiplying. \(\begin{array}{l} \\ \\ (2x-7)(7x+1) \\ 14{x}^{2}+2x-49x-7 \\ 14{x}^{2}-47x-7\ ✓\end{array}\)

Factor completely: \(8{a}^{2}-3a-5\).

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\((a-1)(8a+5)\)
Factor completely: \(6{b}^{2}-b-15\).

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\((2b+3)(3b-5)\)

Factor completely: \(18{n}^{2}-37n+15\).

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The trinomial is already in descending order.	\(18{n}^{2}-37n+15\)
Find the factors of the first term.
Find the factors of the last term. Consider the signs. Since 15 is positive and the coefficient of the middle term is negative, we use the negative facotrs.

Consider all the combinations of factors.

The correct factors are those whose product is the original trinomial.	\((2n-3)(9n-5)\)
Check by multiplying. \(\begin{array}{l} \\ \\ (2n-3)(9n-5) \\ 18{n}^{2}-10n-27n+15 \\ 18{n}^{2}-37n+15\ ✓\end{array}\)

Factor completely: \(18{x}^{2}-3x-10\).

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\((3x+2)(6x-5)\)
Factor completely: \(30{y}^{2}-53y-21\).

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\((3y+1)(10y-21)\)

Factor completely: \(10{y}^{4}+55{y}^{3}+60{y}^{2}\).

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	\(10{y}^{4}+55{y}^{3}+60{y}^{2}\)
Notice the greatest common factor, and factor it first.	\(15{y}^{2}(2{y}^{2}+11y+12)\)
Factor the trinomial.

Consider all the combinations.

The correct factors are those whose product is the original trinomial. Remember to include the factor 5y².	\(5{y}^{2}(y+4)(2y+3)\)
Check by multiplying. \(\begin{array}{l} \\ \\ \\ 5{y}^{2}(y+4)(2y+3) \\ 5{y}^{2}(2{y}^{2}+8y+3y+12) \\ 10{y}^{4}+55{y}^{3}+60{y}^{2}\ ✓\end{array}\)

Factor completely: \(15{n}^{3}-85{n}^{2}+100n\).

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\(5n(n-4)(3n-5)\)
Factor completely: \(56{q}^{3}+320{q}^{2}-96q\).

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\(8q(q+6)(7q-2)\)
Factor: \(6{x}^{2}+7x+2\).
Factor: \(6{x}^{2}+13x+2\).

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\((x+2)(6x+1)\)
Factor: \(4{y}^{2}+8y+3\).

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\((2y+1)(2y+3)\)

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

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

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

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

Split the middle term using 7u and −24u.	\(\begin{array}{l} \\ \\ \\ 8{u}^{2}-\underset{\text{↙}}{17}\underset{\text{↘}}{u}-21 \\ \underset{⎵}{8{u}^{2}+7u}\ \underset{⎵}{-24u-21}\end{array}\)
Factor by grouping.	\(\begin{array}{l} \\ u(8u+7)-3(8u+7) \\ (8u+7)(u-3)\end{array}\)
Check by multiplying. \(\begin{array}{l} \\ \\ (8u+7)(u-3) \\ 8{u}^{2}-24u+7u-21 \\ 8{u}^{2}-17u-21\ ✓\end{array}\)

Factor: \(20{h}^{2}+13h-15\).

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\((4h+5)(5h-3)\)
Factor: \(6{g}^{2}+19g-20\).

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\((g+4)(6g-5)\)
Factor: \(2{x}^{2}+6x+5\).

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Is there a greatest common factor? No.
Find \(a⋅c\). \(a⋅c\)
\(2(5)\)
\(10\)
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.
Factor: \(10{t}^{2}+19t-15\).

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\((2t+5)(5t-3)\)
Factor: \(3{u}^{2}+8u+5\).

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\((u+1)(3u+5)\)

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

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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.
\(5(2{y}^{2}-2y-4y+14)\)
\(5(2{y}^{2}-11y+14)\)
\(10{y}^{2}-55y+70✓\)

Symbols used here

\pm

plus or minus
Both signs at once: x = 3 ± 2 means 5 and 1.

\neq

not equal
The two sides are different.

\leq,\ \geq

less/greater than or equal
Inequalities that allow equality; < and > exclude it.

\sqrt{x},\ \sqrt[n]{x}

square root, n-th root
The non-negative number whose square (n-th power) is x.

|x|

absolute value / modulus
Distance from zero: |−3| = 3. For a complex number, distance from the origin.

imaginary unit
i² = −1.

\log_b x,\ \ln x

logarithm, natural log
The exponent b must be raised to for x; ln uses base e.

\mathbb{N},\ \mathbb{Z},\ \mathbb{Q},\ \mathbb{R},\ \mathbb{C}

number sets
Naturals, integers, rationals, reals, complex numbers.

How to: Factor Trinomials of the Form ax

Recognize a preliminary strategy to factor polynomials completely
Factor trinomials of the form
Factor trinomials using trial and error
Factor trinomials using the ‘ac’ method
Is there a greatest common factor?
Factor it out.
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.

Questions people ask

What does it mean to solve an equation?

To find every value of the unknown that makes both sides equal. Each step is an operation applied to both sides that keeps the solution set the same, until the unknown stands alone.

Why do I sometimes get two answers?

A quadratic can cross the axis twice, so it can have two solutions. A degree-n polynomial has up to n. The graph shows where each one comes from.

How do I know whether to factor or use the quadratic formula?

Try factoring for a few seconds: look for two numbers that multiply to a·c and add to b. If nothing obvious appears, the discriminant b² − 4ac tells you how many real roots there are, and the formula finds them without guessing.

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Parts of this page are adapted from OpenStax Elementary Algebra 2e (CC BY-NC-SA 4.0). Condensed and re-explained here; errors are ours.

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