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Factor Trinomials of the Form x

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 start with the product and end up with the factors. Let’s look at an example of multiplying binomials to refresh your memory.

To factor the trinomial means to start with the product, \({x}^{2}+5x+6\), and end with the factors, \((x+2)(x+3)\). You need to think about where each of the terms in the trinomial came from.

The first term came from multiplying the first term in each binomial. So to get \({x}^{2}\) in the product, each binomial must start with an x.

\[\begin{array}{l}{x}^{2}+5x+6 \\ (x\ )(x\ )\end{array}\]

The last term in the trinomial came from multiplying the last term in each binomial. So the last terms must multiply to 6.

What two numbers multiply to 6?

The factors of 6 could be 1 and 6, or 2 and 3. How do you know which pair to use?

Consider the middle term. It came from adding the outer and inner terms.

Factors of \(6\)	Sum of factors
\(1,6\)	\(1+6=7\)
\(2,3\)	\(2+3=5\)

\[\begin{array}{llll}{x}^{2}+5x+6 & & & \text{product} \\ (x+2)(x+3) & & & \text{factors}\end{array}\]\[\begin{array}{llll}{x}^{2}+5x+6 & & & {x}^{2}+bx+c \\ (x+2)(x+3) & & & (x+m)(x+n)\end{array}\]

How to Factor Trinomials of the Form

Try it.

Factor: \({x}^{2}+7x+12\).

Solution

Example

Try it.

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

Solution

Notice that the variable is u, so the factors will have first terms u.

\(\begin{array}{llll} & & & {u}^{2}+11u+24 \\ \text{Write the factors as two binomials with first terms}\ u. & & & (u\ )(u\ )\end{array}\)

Find two numbers that: multiply to 24 and add to 11.

Factors of \(24\)	Sum of factors
\(\ 1,24\)	\(1+24=25\)
\(\ 2,12\)	\(2+12=14\)
\(3,8\)	\(\ 3+8=11\text{*}\)
\(4,6\)	\(\ 4+6=10\)

\(\begin{array}{llll}\text{Use 3 and 8 as the last terms of the binomials.} & & & \ (u+3)(u+8) \\ \\ \text{Check.} & & & \\ \\ \\ (u+3)(u+8) & & & \\ {u}^{2}+3u+8u+24 & & & \\ {u}^{2}+11u+24\ ✓ & & & \end{array}\)

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

Factor Trinomials of the Form

Sometimes you’ll need to factor trinomials of the form \({x}^{2}+bxy+c{y}^{2}\) with two variables, such as \({x}^{2}+12xy+36{y}^{2}.\) The first term, \({x}^{2}\), is the product of the first terms of the binomial factors, \(x\cdot x\). The \({y}^{2}\) in the last term means that the second terms of the binomial factors must each contain y. To get the coefficients b and c, you use the same process summarized in the previous objective.

Example

Try it.

Factor: \({x}^{2}+12xy+36{y}^{2}\).

Solution

	\({x}^{2}+12xy+36{y}^{2}\)
Note that the first terms are x, last terms contain y.	\((x_y)(x_y)\)

Find the numbers that multiply to 36 and add to 12.

Factors of \(36\)	Sum of factors
1, 36	\(1+36=37\)
2, 18	\(2+18=20\)
3, 12	\(3+12=15\)
4, 9	\(4+9=13\)
6, 6	\(6+6=12\text{*}\)

Use 6 and 6 as the coefficients of the last terms.	\((x+6y)(x+6y)\)
Check your answer. \(\begin{array}{l} \\ \\ (x+6y)(x+6y) \\ {x}^{2}+6xy+6xy+36{y}^{2} \\ {x}^{2}+12xy+36{y}^{2}✓\end{array}\)

Example

Try it.

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

Solution

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.

\(\begin{array}{llll} & & & {r}^{2}-8rs-9{s}^{2} \\ \text{Note that the first terms are}\ r,\ \text{last terms contain}\ s. & & & (r_s)(r_s)\end{array}\)

Find the numbers that multiply to \(-9\) and add to \(-8\).

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

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

Example

Try it.

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

Solution

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

Note there are no factor pairs that give us \(-9\) as a sum. The trinomial is prime.

Key Concepts

Factor trinomials of the form \({x}^{2}+bx+c\)
1. Write the factors as two binomials with first terms x: \((x\ )(x\ )\).
2. Find two numbers m and n that
  Multiply to c, \(m\cdot n=c\)
  Add to b, \(m+n=b\)
3. Use m and n as the last terms of the factors: \((x+m)(x+n)\).
4. Check by multiplying the factors.

Factor Trinomials of the Form x

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.

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

Solution

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

Try it.

\({y}^{2}+8y+7\)

Try it.

\({m}^{2}+12m+11\)

Solution

\((m+1)(m+11)\)

Try it.

\({b}^{2}+14b+13\)

Try it.

\({a}^{2}+9a+20\)

Solution

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

Try it.

\({m}^{2}+7m+12\)

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.

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

Solution

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

Try it.

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

Try it.

\({y}^{2}-6y-7\)

Solution

\((y+1)(y-7)\)

Try it.

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

Try it.

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

Solution

\((x-4)(x+3)\)

Try it.

\({r}^{2}-2r-8\)

Try it.

\({a}^{2}-3a-28\)

Solution

\((a-7)(a+4)\)

Try it.

\({b}^{2}-13b-30\)

Try it.

\({w}^{2}-5w-36\)

Solution

\((w-9)(w+4)\)

Try it.

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

Try it.

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

Solution

prime

Try it.

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

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.

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

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

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.

\({p}^{2}+3pq+2{q}^{2}\)

Solution

\((p+q)(p+2q)\)

Try it.

\({m}^{2}+6mn+5{n}^{2}\)

Try it.

\({r}^{2}+15rs+36{s}^{2}\)

Solution

\((r+3s)(r+12s)\)

Try it.

\({u}^{2}+10uv+24{v}^{2}\)

Try it.

\({m}^{2}-12mn+20{n}^{2}\)

Solution

\((m-2n)(m-10n)\)

Try it.

\({p}^{2}-16pq+63{q}^{2}\)

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.

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

Mixed Practice

In the following exercises, factor each expression.

Try it.

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

Solution

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

Try it.

\({w}^{2}+4w-32\)

Try it.

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

Solution

\((x+2)(x-16)\)

Try it.

\({y}^{2}+41y+40\)

Try it.

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

Solution

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

Try it.

\({x}^{2}-16xy+64{y}^{2}\)

Try it.

\({k}^{2}+34k+120\)

Solution

\((k+4)(k+30)\)

Try it.

\({m}^{2}+29m+120\)

Try it.

\({y}^{2}+10y+15\)

Solution

prime

Try it.

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

Try it.

\({m}^{2}+mn-56{n}^{2}\)

Solution

\((m+8n)(m-7n)\)

Try it.

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

Try it.

\({u}^{2}-17uv+30{v}^{2}\)

Solution

\((u-15v)(u-2v)\)

Try it.

\({m}^{2}-31mn+30{n}^{2}\)

Try it.

\({c}^{2}-8cd+26{d}^{2}\)

Solution

prime

Try it.

\({r}^{2}+11rs+36{s}^{2}\)

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.

Multiply: \((x+4)(x+5).\)
If you missed this problem, review .

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\({x}^{2}+9x+20\)
Simplify: ⓐ \(-9+(-6)\) ⓑ \(-9+6.\)
If you missed this problem, review .

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ⓐ −15 ⓑ −3
Simplify: ⓐ \(-9(6)\) ⓑ \(-9(-6).\)
If you missed this problem, review .

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ⓐ −54 ⓑ 54
Simplify: ⓐ \(|-5|\) ⓑ \(|3|.\)
If you missed this problem, review .

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Factor: \({x}^{2}+7x+12\).
Factor: \({x}^{2}+6x+8\).

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\((x+2)(x+4)\)
Factor: \({y}^{2}+8y+15\).

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\((y+3)(y+5)\)
Factor: \({u}^{2}+11u+24\).

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Notice that the variable is u, so the factors will have first terms u.
\(\begin{array}{llll} & & & {u}^{2}+11u+24 \\ \text{Write the factors as two binomials with first terms}\ u. & & & (u\ )(u\ )\end{array}\)
Find two numbers that: multiply to 24 and add to 11.
Factors of \(24\) Sum of factors
\(\ 1,24\) \(1+24=25\)
\(\ 2,12\) \(2+12=14\)
\(3,8\) \(\ 3+8=11\text{*}\)
\(4,6\) \(\ 4+6=10\)
\(\begin{array}{llll}\text{Use 3 and 8 as the last terms of the binomials.} & & & \ (u+3)(u+8) \\ \\ \text{Check.} & & & \\ \\ \\ (u+3)(u+8) & & & \\ {u}^{2}+3u+8u+24 & & & \\ {u}^{2}+11u+24\ ✓ & & & \end{array}\)
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: \({y}^{2}+17y+60\).

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\(\begin{array}{llll} & & & {y}^{2}+17y+60 \\ \text{Write the factors as two binomials with first terms}\ y. & & & (y\ )(y\ )\end{array}\)
Find two numbers that multiply to 60 and add to 17.
Factors of \(60\) Sum of factors
\(1,60\) \(1+60=61\)
\(2,30\) \(2+30=32\)
\(3,20\) \(3+20=23\)
\(4,15\) \(4+15=19\)
\(5,12\) \(\ 5+12=17\text{*}\)
\(6,10\) \(6+10=16\)
\(\begin{array}{llll}\text{Use}\ 5\ \text{and}\ 12\ \text{as the last terms.} & & & \ (y+5)(y+12) \\ \\ \text{Check.} & & & \\ \\ \\ (y+5)(y+12) & & & \\ ({y}^{2}+12y+5y+60) & & & \\ ({y}^{2}+17y+60)✓ & & & \end{array}\)
Factor: \({x}^{2}+19x+60\).

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\((x+4)(x+15)\)
Factor: \({v}^{2}+23v+60\).

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\((v+3)(v+20)\)
Factor: \({t}^{2}-11t+28\).

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Again, with the positive last term, 28, and the negative middle term, \(-11t\), we need two negative factors. Find two numbers that multiply 28 and add to \(-11\).
\(\begin{array}{llll} & & & {t}^{2}-11t+28 \\ \\ \text{Write the factors as two binomials with first terms}\ t. & & & (t\ )(t\ )\end{array}\)
Find two numbers that: multiply to 28 and add to \(-11\).
Factors of \(28\) Sum of factors
\(\ -1,-28\) \(-1+(-28)=-29\)
\(\ -2,-14\) \(-2+(-14)=-16\)
\(-4,-7\) \(\ -4+(-7)=-11\text{*}\)
\(\begin{array}{llll}\text{Use}\ -4,-7\ \text{as the last terms of the binomials.} & & & \ (t-4)(t-7) \\ \text{Check.} & & & \\ \\ \\ (t-4)(t-7) & & & \\ {t}^{2}-7t-4t+28 & & & \\ {t}^{2}-11t+28\ ✓ & & & \end{array}\)
Factor: \({u}^{2}-9u+18\).

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

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\((y-7)(y-9)\)

Factor: \({z}^{2}+4z-5\).

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To get a negative last term, multiply one positive and one negative. We need factors of \(-5\) that add to positive 4.

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

Notice: We listed both \(1,-5\ \text{and}\ -1,5\) to make sure we got the sign of the middle term correct.

	\({z}^{2}+4z-5\)
Factors will be two binomials with first terms z.	\((z\ )(z\ )\)
Use −1, 5 as the last terms of the binomials.	\((z-1)(z+5)\)
Check. \(\begin{array}{l} \\ (z-1)(z+5) \\ {z}^{2}+5z-1z-5 \\ {z}^{2}+4z-5\ ✓\end{array}\)

Factor: \({h}^{2}+4h-12\).

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\((h-2)(h+6)\)
Factor: \({k}^{2}+k-20\).

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\((k-4)(k+5)\)

Factor: \({z}^{2}-4z-5\).

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This time, we need factors of \(-5\) that add to \(-4\).

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

	\({z}^{2}-4z-5\)
Factors will be two binomials with first terms z.	\((z\ )(z\ )\)
Use 1, −5 as the last terms of the binomials.	\((z+1)(z-5)\)
Check. \(\begin{array}{l} \\ (z+1)(z-5) \\ {z}^{2}-5z+1z-5 \\ {z}^{2}-4z-5\ ✓\end{array}\)

Notice that the factors of \({z}^{2}-4z-5\) are very similar to the factors of \({z}^{2}+4z-5\). It is very important to make sure you choose the factor pair that results in the correct sign of the middle term.

Factor: \({x}^{2}-4x-12\).

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\((x+2)(x-6)\)
Factor: \({y}^{2}-y-20\).

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\((y+4)(y-5)\)
Factor: \({q}^{2}-2q-15\).

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\(\begin{array}{llll} & & & {q}^{2}-2q-15 \\ \text{Factors will be two binomials with first terms}\ q. & & & (q\ )(q\ ) \\ \text{You can use}\ 3,-5\ \text{as the last terms of the} & & & (q+3)(q-5) \\ \text{binomials.} & & & \end{array}\)
Factors of \(-15\) Sum of factors
\(\ 1,-15\) \(1+(-15)=-14\)
\(-1,15\) \(\ -1+15=14\)
\(\ 3,-5\) \(\ 3+(-5)=-2\text{*}\)
\(-3,5\) \(\ -3+5=2\)
Check.
\(\ (q+3)(q-5)\)
\(\ {q}^{2}-5q+3q-15\)
\(\ {q}^{2}-2q-15\ ✓\)
Factor: \({r}^{2}-3r-40\).

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\((r+5)(r-8)\)
Factor: \({s}^{2}-3s-10\).

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\((s+2)(s-5)\)
Factor: \({y}^{2}-6y+15\).

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\({y}^{2}-6y+15\)
Factors will be two binomials with first terms y. \((y\ )(y\ )\)

Factors of 15 Sum of factors
\(-1,-15\) \(-1+(-15)=-16\)
\(-3,-5\) \(\ -3+(-5)=-8\)
As shown in the table, none of the factors add to \(-6\); therefore, the expression is prime.
Factor: \({m}^{2}+4m+18\).

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prime
Factor: \({n}^{2}-10n+12\).

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prime
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. \((x\ )(x\ )\)

As shown in the table, you can use \(-6,8\) as the last terms of the binomials.
\[\ (x-6)(x+8)\]
Factors of \(-48\) Sum of factors
\(-1,48\) \(-1+48=47\)
\(-2,24\)
\(-3,16\)
\(-4,12\)
\(-6,8\) \(-2+24=22\)
\(-3+16=13\)
\(-4+12=8\)
\(\ -6+8=2\)
Check.
\(\ (x-6)(x+8)\)
\(\ {x}^{2}-6x+8x-48\)
\(\ {x}^{2}+2x-48\ ✓\)
Factor: \(9m+{m}^{2}+18\).

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

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\((n-3)(n-4)\)
Factor: \({x}^{2}+12xy+36{y}^{2}\).

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\({x}^{2}+12xy+36{y}^{2}\)
Note that the first terms are x, last terms contain y. \((x_y)(x_y)\)

Find the numbers that multiply to 36 and add to 12.
Factors of \(36\) Sum of factors
1, 36 \(1+36=37\)
2, 18 \(2+18=20\)
3, 12 \(3+12=15\)
4, 9 \(4+9=13\)
6, 6 \(6+6=12\text{*}\)
Use 6 and 6 as the coefficients of the last terms. \((x+6y)(x+6y)\)
Check your answer.
\(\begin{array}{l} \\ \\ (x+6y)(x+6y) \\ {x}^{2}+6xy+6xy+36{y}^{2} \\ {x}^{2}+12xy+36{y}^{2}✓\end{array}\)
Factor: \({u}^{2}+11uv+28{v}^{2}\).

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\((u+4v)(u+7v)\)
Factor: \({x}^{2}+13xy+42{y}^{2}\).

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\((x+6y)(x+7y)\)
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.
\(\begin{array}{llll} & & & {r}^{2}-8rs-9{s}^{2} \\ \text{Note that the first terms are}\ r,\ \text{last terms contain}\ s. & & & (r_s)(r_s)\end{array}\)
Find the numbers that multiply to \(-9\) and add to \(-8\).
Factors of \(-9\) Sum of factors
\(\ 1,-9\) \(\ 1+(-9)=-8\text{*}\)
\(-1,9\) \(\ -1+9=8\)
\(\ 3,-3\) \(3+(-3)=0\)
\(\begin{array}{llll}\text{Use}\ 1,-9\ \text{as coefficients of the last terms.} & & & \ (r+s)(r-9s) \\ \text{Check your answer.} & & & \\ \\ \\ (r-9s)(r+s) & & & \\ {r}^{2}+rs-9rs-9{s}^{2} & & & \\ {r}^{2}-8rs-9{s}^{2}\ ✓ & & & \end{array}\)
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: \({u}^{2}-9uv-12{v}^{2}\).

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

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 x

Factor trinomials of the form
Factor trinomials of the form
Write the factors as two binomials with first terms
Find two numbers
Use
Check by multiplying the factors.
the first term,
the positive last term is the product of the two last terms

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.

Ku day inaad ku

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.

In ka badan Algebra

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

Factors of \(60\)	Sum of factors
\(1,60\)	\(1+60=61\)
\(2,30\)	\(2+30=32\)
\(3,20\)	\(3+20=23\)
\(4,15\)	\(4+15=19\)
\(5,12\)	\(\ 5+12=17\text{*}\)
\(6,10\)	\(6+10=16\)

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

Factors of \(-15\)	Sum of factors
\(\ 1,-15\)	\(1+(-15)=-14\)
\(-1,15\)	\(\ -1+15=14\)
\(\ 3,-5\)	\(\ 3+(-5)=-2\text{*}\)
\(-3,5\)	\(\ -3+5=2\)

	\({y}^{2}-6y+15\)
Factors will be two binomials with first terms y.	\((y\ )(y\ )\)

Factors of 15	Sum of factors
\(-1,-15\)	\(-1+(-15)=-16\)
\(-3,-5\)	\(\ -3+(-5)=-8\)

	\(2x+{x}^{2}-48\)
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
\(-1,48\)	\(-1+48=47\)
\(-2,24\) \(-3,16\) \(-4,12\) \(-6,8\)	\(-2+24=22\) \(-3+16=13\) \(-4+12=8\) \(\ -6+8=2\)