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Factor Trinomials

Factor trinomials of the form

Factor Trinomials of the Form

You have already learned how to multiply binomials using FOIL. Now you’ll need to “undo” this multiplication. To factor the trinomial means to start with the product, and end with the factors.

To figure out how we would factor a trinomial of the form \({x}^{2}+bx+c,\) such as \({x}^{2}+5x+6\) and factor it to \((x+2)(x+3),\) let’s start with two general binomials of the form \((x+m)\) and \((x+n).\)


Foil to find the product.
Factor the GCF from the middle terms.
Our trinomial is of the form \({x}^{2}+bx+c.\)

This tells us that to factor a trinomial of the form \({x}^{2}+bx+c,\) we need two factors \((x+m)\) and \((x+n)\) where the two numbers m and n multiply to c and add to b.

How to Factor a Trinomial of the form

Try it.

Factor: \({x}^{2}+11x+24.\)

Solution

Let’s summarize the steps we used to find the factors.

In the first example, all terms in the trinomial were positive. What happens when there are negative terms? Well, it depends which term is negative. Let’s look first at trinomials with only the middle term negative.

How do you get a positive product and a negative sum? We use two negative numbers.

Example

Try it.

Factor: \({y}^{2}-11y+28.\)

Solution

Again, with the positive last term, 28, and the negative middle term, \(-11y,\) we need two negative factors. Find two numbers that multiply 28 and add to \(-11.\)

	\({y}^{2}-11y+28\)
Write the factors as two binomials with first terms \(y.\)	\((y\ )(y\ )\)
Find two numbers that: multiply to 28 and add to −11.

Factors of \(28\)	Sum of factors
\(\text{-}1,-28\) \(\text{-}2,-14\) \(\text{-}4,-7\)	\(\text{-}1+(\text{-}28)=-29\) \(\text{-}2+(\text{-}14)=-16\) \(\ \text{-}4+(\text{-}7)={-11}^{*}\)

Use \(-4,-7\) as the last terms of the binomials.	\((y-4)(y-7)\)
Check: \(\ \begin{array}{l} \\ \\ (y-4)(y-7) \\ {y}^{2}-7y-4y+28 \\ {y}^{2}-11y+28✓\end{array}\)

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

Factor Trinomials of the form

Our next step is 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 we’ve used so far. Let’s do an example to see how this works.

Example

Try it.

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

Solution

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

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

Check:
\(\ \begin{array}{l} \\ \\ 4x(x-1)(x+5) \\ 4x({x}^{2}+5x-x-5) \\ 4x({x}^{2}+4x-5) \\ 4{x}^{3}+16{x}^{2}-20x✓\end{array}\)

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

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

How to Factor a Trinomial Using Trial and Error

Try it.

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

Solution

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

Factor Trinomials of the Form

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 using the ‘ac’ method: \(6{x}^{2}+7x+2.\)

Solution

The “ac” method is summarized here.

Don’t forget to look for a common factor!

Example

Try it.

Factor using the ‘ac’ method: \(10{y}^{2}-55y+70.\)

Solution

Is there a greatest common factor?
Yes. The GCF is 5.
Factor it.
The trinomial inside the parentheses has a leading coefficient that is not 1.
Find the product \(ac.\)	\(\ ac=28\)
Find two numbers that multiply to \(ac\)	\(\ (-4)(-7)=28\)
and add to b.	\(-4+(-7)=-11\)
Split the middle term.

Factor the trinomial by grouping.

Check by multiplying all three factors. \(\ \begin{array}{l}5(y-2)(2y-7) \\ 5(2{y}^{2}-7y-4y+14) \\ 5(2{y}^{2}-11y+14) \\ 10{y}^{2}-55y+70✓\end{array}\)

Factor Using Substitution

Sometimes a trinomial does not appear to be in the \(a{x}^{2}+bx+c\) form. However, we can often make a thoughtful substitution that will allow us to make it fit the \(a{x}^{2}+bx+c\) form. This is called factoring by substitution. It is standard to use u for the substitution.

In the \(a{x}^{2}+bx+c,\) the middle term has a variable, x, and its square, \({x}^{2},\) is the variable part of the first term. Look for this relationship as you try to find a substitution.

Example

Try it.

Factor by substitution: \({x}^{4}-4{x}^{2}-5.\)

Solution

The variable part of the middle term is \({x}^{2}\) and its square, \({x}^{4},\) is the variable part of the first term. (We know \({({x}^{2})}^{2}={x}^{4}).\) If we let \(u={x}^{2},\) we can put our trinomial in the \(a{x}^{2}+bx+c\) form we need to factor it.


Rewrite the trinomial to prepare for the substitution.
Let \(u={x}^{2}\) and substitute.
Factor the trinomial.
Replace u with \({x}^{2}.\)
Check: \(\ \begin{array}{l}({x}^{2}+1)({x}^{2}-5) \\ {x}^{4}-5{x}^{2}+{x}^{2}-5 \\ {x}^{4}-4{x}^{2}-5✓\end{array}\)

Sometimes the expression to be substituted is not a monomial.

Example

Try it.

Factor by substitution: \({(x-2)}^{2}+7(x-2)+12\)

Solution

The binomial in the middle term, \((x-2)\) is squared in the first term. If we let \(u=x-2\) and substitute, our trinomial will be in \(a{x}^{2}+bx+c\) form.


Rewrite the trinomial to prepare for the substitution.
Let \(u=x-2\) and substitute.
Factor the trinomial.
Replace u with \(x-2.\)
Simplify inside the parentheses.

This could also be factored by first multiplying out the \({(x-2)}^{2}\) and the \(7(x-2)\) and then combining like terms and then factoring. Most students prefer the substitution method.

Key Concepts

How to factor trinomials of the form \({x}^{2}+bx+c.\)
1. Write the factors as two binomials with first terms x. \(\ \begin{array}{l}{x}^{2}+bx+c \\ (x\ )(x\ )\end{array}\)
2. Find two numbers m and n that
  \(\begin{array}{lll}\text{multiply to} & & c,m\cdot n=c \\ \text{add to} & & b,m+n=b\end{array}\)
3. Use m and n as the last terms of the factors. \(\ (x+m)(x+n)\)
4. Check by multiplying the factors.
Strategy for Factoring Trinomials of the Form \({x}^{2}+bx+c\): When we factor a trinomial, we look at the signs of its terms first to determine the signs of the binomial factors.
\(\begin{array}{l}\begin{array}{l}{x}^{2}+bx+c \\ (x+m)(x+n)\end{array} \\ \text{When}\ \text{c}\ \text{is positive,}\ \text{m}\ \text{and}\ \text{n}\ \text{have the same sign.} \\ b\ \text{positive}\ b\ \text{negative} \\ m,n\ \text{positive}\ m,n\ \text{negative} \\ {x}^{2}+5x+6\ {x}^{2}-6x+8 \\ (x+2)(x+3)\ (x-4)(x-2) \\ \text{same signs}\ \text{same signs} \\ \text{When}\ \text{c}\ \text{is negative,}\ \text{m}\ \text{and}\ \text{n}\ \text{have opposite signs.} \\ {x}^{2}+x-12\ {x}^{2}-2x-15 \\ (x+4)(x-3)\ (x-5)(x+3) \\ \text{opposite signs}\ \text{opposite signs}\end{array}\)
Notice that, in the case when m and n have opposite signs, the sign of the one with the larger absolute value matches the sign of b.
How to factor trinomials of the form \(a{x}^{2}+bx+c\) using trial and error.
1. Write the trinomial in descending order of degrees as needed.
2. Factor any GCF.
3. Find all the factor pairs of the first term.
4. Find all the factor pairs of the third term.
5. Test all the possible combinations of the factors until the correct product is found.
6. Check by multiplying.
How to factor trinomials of the form \(a{x}^{2}+bx+c\) using the “ac” method.
1. Factor any GCF.
2. Find the product ac.
3. Find two numbers m and n that:
  \(\begin{array}{llllll}\text{Multiply to}\ ac. & & & & & m\cdot n=a\cdot c \\ \text{Add to}\ b. & & & & & m+n=b \\ & & & & & a{x}^{2}+bx+c\end{array}\)
4. Split the middle term using m and n. \(\ a{x}^{2}+mx+nx+c\)
5. Factor by grouping.
6. Check by multiplying the factors.

Factor Trinomials

Factor Trinomials of the Form \({x}^{2}+bx+c\)

In the following exercises, factor each trinomial of the form \({x}^{2}+bx+c.\)

Try it.

\({p}^{2}+11p+30\)

Solution

\((p+5)(p+6)\)

Try it.

\({w}^{2}+10w+21\)

Try it.

\({n}^{2}+19n+48\)

Solution

\((n+3)(n+16)\)

Try it.

\({b}^{2}+14b+48\)

Try it.

\({a}^{2}+25a+100\)

Solution

\((a+5)(a+20)\)

Try it.

\({u}^{2}+101u+100\)

Try it.

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

Solution

\((x-2)(x-6)\)

Try it.

\({q}^{2}-13q+36\)

Try it.

\({y}^{2}-18y+45\)

Solution

\((y-3)(y-15)\)

Try it.

\({m}^{2}-13m+30\)

Try it.

\({x}^{2}-8x+7\)

Solution

\((x-1)(x-7)\)

Try it.

\({y}^{2}-5y+6\)

Try it.

\(5p-6+{p}^{2}\)

Solution

\((p-1)(p+6)\)

Try it.

\(6n-7+{n}^{2}\)

Try it.

\(8-6x+{x}^{2}\)

Solution

\((x-4)(x-2)\)

Try it.

\(7x+{x}^{2}+6\)

Try it.

\({x}^{2}-12-11x\)

Solution

\((x-12)(x+1)\)

Try it.

\(-11-10x+{x}^{2}\)

In the following exercises, factor each trinomial of the form \({x}^{2}+bxy+c{y}^{2}.\) If the trinomial cannot be factored, answer “Prime.”

Try it.

\({x}^{2}-2xy-80{y}^{2}\)

Solution

\((x+8y)(x-10y)\)

Try it.

\({p}^{2}-8pq-65{q}^{2}\)

Try it.

\({m}^{2}-64mn-65{n}^{2}\)

Solution

\((m+n)(m-65n)\)

Try it.

\({p}^{2}-2pq-35{q}^{2}\)

Try it.

\({a}^{2}+5ab-24{b}^{2}\)

Solution

\((a+8b)(a-3b)\)

Try it.

\({r}^{2}+3rs-28{s}^{2}\)

Try it.

\({x}^{2}-3xy-14{y}^{2}\)

Solution

Prime

Try it.

\({u}^{2}-8uv-24{v}^{2}\)

Try it.

\({m}^{2}-5mn+30{n}^{2}\)

Solution

Prime

Try it.

\({c}^{2}-7cd+18{d}^{2}\)

Factor Trinomials of the Form \(a{x}^{2}+bx+c\) Using Trial and Error

In the following exercises, factor completely using trial and error.

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.

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

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.

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

Solution

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

Try it.

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

Try it.

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

Solution

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

Try it.

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

Try it.

\(4{a}^{2}+17ab-15{b}^{2}\)

Solution

\((4a-3b)(a+5b)\)

Try it.

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

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 of the Form \(a{x}^{2}+bx+c\) using the ‘ac’ Method

In the following exercises, factor using the ‘ac’ method.

Try it.

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

Solution

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

Try it.

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

Try it.

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

Solution

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

Try it.

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

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.

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

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.

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

Factor Using Substitution

In the following exercises, factor using substitution.

Try it.

\({x}^{4}-6{x}^{2}-7\)

Solution

\(({x}^{2}+1)({x}^{2}-7)\)

Try it.

\({x}^{4}+2{x}^{2}-8\)

Try it.

\({x}^{4}-3{x}^{2}-28\)

Solution

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

Try it.

\({x}^{4}-13{x}^{2}-30\)

Try it.

\({(x-3)}^{2}-5(x-3)-36\)

Solution

\((x-12)(x+1)\)

Try it.

\({(x-2)}^{2}-3(x-2)-54\)

Try it.

\({(3y-2)}^{2}-(3y-2)-2\)

Solution

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

Try it.

\({(5y-1)}^{2}-3(5y-1)-18\)

Mixed Practice

In the following exercises, factor each expression using any method.

Try it.

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

Solution

\((u-6)(u-6)\)

Try it.

\({x}^{2}-14x-32\)

Try it.

\({r}^{2}-20rs+64{s}^{2}\)

Solution

\((r-4s)(r-16s)\)

Try it.

\({q}^{2}-29qr-96{r}^{2}\)

Try it.

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

Solution

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

Try it.

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

Try it.

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

Solution

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

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.

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

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.

\({x}^{4}-4{x}^{2}-12\)

Solution

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

Try it.

\({x}^{4}-7{x}^{2}-8\)

Try it.

\({(x+3)}^{2}-9(x+3)-36\)

Solution

\((x-9)(x+6)\)

Try it.

\({(x+2)}^{2}-25(x+2)-54\)

Condensed — the full section is in OpenStax Intermediate 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 all the factors of 72.
If you missed this problem, review .

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1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36, 72
Find the product: \((3y+4)(2y+5).\)
If you missed this problem, review .

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\(6{y}^{2}+23y+20\)
Simplify: \(-9(6);\) \(-9(-6).\)
If you missed this problem, review .

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−54, 54
Factor: \({x}^{2}+11x+24.\)
Factor: \({q}^{2}+10q+24.\)

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\((q+4)(q+6)\)
Factor: \({t}^{2}+14t+24.\)

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\((t+2)(t+12)\)
Factor: \({y}^{2}-11y+28.\)

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Again, with the positive last term, 28, and the negative middle term, \(-11y,\) we need two negative factors. Find two numbers that multiply 28 and add to \(-11.\)

\({y}^{2}-11y+28\)
Write the factors as two binomials with first terms \(y.\) \((y\ )(y\ )\)
Find two numbers that: multiply to 28 and add to −11.

Factors of \(28\) Sum of factors
\(\text{-}1,-28\)

\(\text{-}2,-14\)

\(\text{-}4,-7\) \(\text{-}1+(\text{-}28)=-29\)

\(\text{-}2+(\text{-}14)=-16\)

\(\ \text{-}4+(\text{-}7)={-11}^{*}\)

Use \(-4,-7\) as the last terms of the binomials. \((y-4)(y-7)\)
Check:
\(\ \begin{array}{l} \\ \\ (y-4)(y-7) \\ {y}^{2}-7y-4y+28 \\ {y}^{2}-11y+28✓\end{array}\)
Factor: \({u}^{2}-9u+18.\)

Avslöja svaret

\((u-3)(u-6)\)
Factor: \({y}^{2}-16y+63.\)

Avslöja svaret

\((y-7)(y-9)\)
Factor: \(2x+{x}^{2}-48.\)

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\(2x+{x}^{2}-48\)
First we put the terms in decreasing degree order. \({x}^{2}+2x-48\)
Factors will be two binomials with first terms \(x.\) \((x\ )(x\ )\)

Factors of \(-48\) Sum of factors
\(-1,48\)
\(-2,24\)
\(-3,16\)
\(-4,12\)
\(-6,8\) \(-1+48=47\)
\(-2+24=22\)
\(-3+16=13\)
\(-4+12=8\)
\(\ -6+8={2}^{*}\)

\(\text{Use}\ -6,8\ \text{as the last terms of the binomials.}\) \((x-6)(x+8)\)
Check:
\(\ \begin{array}{l} \\ \\ (x-6)(x+8) \\ {x}^{2}-6q+8q-48 \\ {x}^{2}+2x-48✓\end{array}\)
Factor: \(9m+{m}^{2}+18.\)

Avslöja svaret

\((m+3)(m+6)\)
Factor: \(-7n+12+{n}^{2}.\)

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\((n-3)(n-4)\)
Factor: \({r}^{2}-8rs-9{s}^{2}.\)

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We need r in the first term of each binomial and s in the second term. The last term of the trinomial is negative, so the factors must have opposite signs.

\({r}^{2}-8rs-9{s}^{2}\)
Note that the first terms are \(r,\) last terms contain \(s.\) \((r\ s)(r\ s)\)
Find the numbers that multiply to −9 and add to −8.

Factors of \(-9\) Sum of factors
\(\ 1,-9\) \(\ -1+9=8\)
\(-1,9\) \(1+(-9)=\text{-}{8}^{*}\)
\(\ 3,-3\) \(3+(-3)=0\)

\(\text{Use}\ 1,-9\ \text{as coefficients of the last terms.}\) \((r+s)(r-9s)\)
Check:
\(\ \begin{array}{l}(r-9s)(r+s) \\ {r}^{2}+rs-9rs-9{s}^{2} \\ {r}^{2}-8rs-9{s}^{2}✓\end{array}\)
Factor: \({a}^{2}-11ab+10{b}^{2}.\)

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\((a-b)(a-10b)\)
Factor: \({m}^{2}-13mn+12{n}^{2}.\)

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\((m-n)(m-12n)\)
Factor: \({u}^{2}-9uv-12{v}^{2}.\)

Avslöja svaret

We need u in the first term of each binomial and v in the second term. The last term of the trinomial is negative, so the factors must have opposite signs.

\({u}^{2}-9uv-12{v}^{2}\)
Note that the first terms are \(u,\) last terms contain \(v.\) \((u\ v)(u\ v)\)
Find the numbers that multiply to −12 and add to −9.

Factors of \(\text{-}12\) Sum of factors
\(\ 1,-12\)
\(-1,12\)
\(\ 2,-6\)
\(-2,6\)
\(\ 3,-4\)
\(-3,4\) \(1+(-12)=-11\)
\(\ -1+12=11\)
\(\ 2+(-6)=-4\)
\(\ -2+6=4\)
\(\ 3+(-4)=-1\)
\(\ -3+4=1\)
Note there are no factor pairs that give us \(-9\) as a sum. The trinomial is prime.
Factor: \({x}^{2}-7xy-10{y}^{2}.\)

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prime
Factor: \({p}^{2}+15pq+20{q}^{2}.\)

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prime

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

Avslöja svaret

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

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

Check:
\(\ \begin{array}{l} \\ \\ 4x(x-1)(x+5) \\ 4x({x}^{2}+5x-x-5) \\ 4x({x}^{2}+4x-5) \\ 4{x}^{3}+16{x}^{2}-20x✓\end{array}\)

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 using trial and error: \(3{y}^{2}+22y+7.\)
Factor completely using trial and error: \(2{a}^{2}+5a+3.\)

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

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

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

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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 using trial and error: \(8{x}^{2}-14x+3.\)

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

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

Factor completely using trial and error: \(18{x}^{2}-37xy+15{y}^{2}.\)

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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 15 is positive and the coefficient of the middle term is negative, we use the negative factors.

Consider all the combinations of factors.

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

Factor completely using trial and error \(18{x}^{2}-3xy-10{y}^{2}.\)

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

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

Factor completely using trial and error: \(-10{y}^{4}-55{y}^{3}-60{y}^{2}.\)

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Notice the greatest common factor, so factor it first.
Factor the trinomial.

Consider all the combinations.

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

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

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

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\(8q(q+6)(7q-2)\)
Factor using the ‘ac’ method: \(6{x}^{2}+7x+2.\)
Factor using the ‘ac’ method: \(6{x}^{2}+13x+2.\)

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\((x+2)(6x+1)\)
Factor using the ‘ac’ method: \(4{y}^{2}+8y+3.\)

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

Factor using the ‘ac’ method: \(10{y}^{2}-55y+70.\)

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Is there a greatest common factor?
Yes. The GCF is 5.
Factor it.
The trinomial inside the parentheses has a leading coefficient that is not 1.
Find the product \(ac.\)	\(\ ac=28\)
Find two numbers that multiply to \(ac\)	\(\ (-4)(-7)=28\)
and add to b.	\(-4+(-7)=-11\)
Split the middle term.

Factor the trinomial by grouping.

Check by multiplying all three factors. \(\ \begin{array}{l}5(y-2)(2y-7) \\ 5(2{y}^{2}-7y-4y+14) \\ 5(2{y}^{2}-11y+14) \\ 10{y}^{2}-55y+70✓\end{array}\)

Factor using the ‘ac’ method: \(16{x}^{2}-32x+12.\)

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\(4(2x-3)(2x-1)\)
Factor using the ‘ac’ method: \(18{w}^{2}-39w+18.\)

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\(3(3w-2)(2w-3)\)

Factor by substitution: \({x}^{4}-4{x}^{2}-5.\)

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Rewrite the trinomial to prepare for the substitution.
Let \(u={x}^{2}\) and substitute.
Factor the trinomial.
Replace u with \({x}^{2}.\)
Check: \(\ \begin{array}{l}({x}^{2}+1)({x}^{2}-5) \\ {x}^{4}-5{x}^{2}+{x}^{2}-5 \\ {x}^{4}-4{x}^{2}-5✓\end{array}\)

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

Factor trinomials of the form
Factor trinomials of the form
Factor trinomials of the form
Factor using substitution
Write the factors as two binomials with first terms
Find two numbers
multiply to
add to

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.

Prova själv

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

Mer information Algebra

Linear equations Quadratic equations Systems of equations Inequalities Factoring Expanding Simplifying expressions Functions and graphs Exponential and logarithmic equations Polynomial equations Absolute value

	\(2x+{x}^{2}-48\)
First we put the terms in decreasing degree order.	\({x}^{2}+2x-48\)
Factors will be two binomials with first terms \(x.\)	\((x\ )(x\ )\)

Factors of \(-48\)	Sum of factors
\(-1,48\) \(-2,24\) \(-3,16\) \(-4,12\) \(-6,8\)	\(-1+48=47\) \(-2+24=22\) \(-3+16=13\) \(-4+12=8\) \(\ -6+8={2}^{*}\)

\(\text{Use}\ -6,8\ \text{as the last terms of the binomials.}\)	\((x-6)(x+8)\)
Check: \(\ \begin{array}{l} \\ \\ (x-6)(x+8) \\ {x}^{2}-6q+8q-48 \\ {x}^{2}+2x-48✓\end{array}\)

	\({r}^{2}-8rs-9{s}^{2}\)
Note that the first terms are \(r,\) last terms contain \(s.\)	\((r\ s)(r\ s)\)
Find the numbers that multiply to −9 and add to −8.

Factors of \(-9\)	Sum of factors
\(\ 1,-9\)	\(\ -1+9=8\)
\(-1,9\)	\(1+(-9)=\text{-}{8}^{*}\)
\(\ 3,-3\)	\(3+(-3)=0\)

\(\text{Use}\ 1,-9\ \text{as coefficients of the last terms.}\)	\((r+s)(r-9s)\)
Check: \(\ \begin{array}{l}(r-9s)(r+s) \\ {r}^{2}+rs-9rs-9{s}^{2} \\ {r}^{2}-8rs-9{s}^{2}✓\end{array}\)

	\({u}^{2}-9uv-12{v}^{2}\)
Note that the first terms are \(u,\) last terms contain \(v.\)	\((u\ v)(u\ v)\)
Find the numbers that multiply to −12 and add to −9.

Factors of \(\text{-}12\)	Sum of factors
\(\ 1,-12\) \(-1,12\) \(\ 2,-6\) \(-2,6\) \(\ 3,-4\) \(-3,4\)	\(1+(-12)=-11\) \(\ -1+12=11\) \(\ 2+(-6)=-4\) \(\ -2+6=4\) \(\ 3+(-4)=-1\) \(\ -3+4=1\)