Factorising Trinomials of the Form <em>x</em><sup>2</sup> + <em>bxy</em> + <em>cy</em><sup>2</sup>
Factorising Trinomials of the Form x2 + bxy + cy2
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, \(x·x\). 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
Factor: \({x}^{2}+12xy+36{y}^{2}\).
Solution
\(\begin{array}{cccc}& & & \phantom{\rule{4em}{0ex}}{x}^{2}+12xy+36{y}^{2}\hfill \\ \begin{array}{c}\text{Note that the first terms are}\phantom{\rule{0.2em}{0ex}}x,\phantom{\rule{0.2em}{0ex}}\text{last terms}\hfill \\ \text{contain}\phantom{\rule{0.2em}{0ex}}y.\hfill \end{array}\hfill & & & \phantom{\rule{4em}{0ex}}\left(x_y\right)\left(x_y\right)\hfill \end{array}\)
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 | \(6+6=12\text{*}\) |
\(\begin{array}{cccc}\text{Use 6 and 6 as the coefficients of the last terms.}\hfill & & & \hfill \phantom{\rule{2em}{0ex}}\left(x+6y\right)\left(x+6y\right)\hfill \\ \text{Check your answer.}\hfill & & & \\ \phantom{\rule{3.5em}{0ex}}\left(x+6y\right)\left(x+6y\right)\hfill & & & \\ \phantom{\rule{2.5em}{0ex}}{x}^{2}+6xy+6xy+36{y}^{2}\hfill & & & \\ \phantom{\rule{2.5em}{0ex}}{x}^{2}+12xy+36{y}^{2}\phantom{\rule{0.2em}{0ex}}✓\hfill & & & \end{array}\)
Example
Factor: \({r}^{2}-8rx-9{s}^{2}\).
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.
\(\begin{array}{cccc}& & & {r}^{2}-8rx-9{s}^{2}\hfill \\ \text{Note that the first terms are}\phantom{\rule{0.2em}{0ex}}r,\phantom{\rule{0.2em}{0ex}}\text{last terms contain}\phantom{\rule{0.2em}{0ex}}s.\hfill & & & \left(r_s\right)\left(r_s\right)\hfill \end{array}\)
Find the numbers that multiply to \(-9\) and add to \(-8\).
| Factors of \(-9\) | Sum of factors |
|---|---|
| \(\phantom{\rule{1em}{0ex}}1,-9\) | \(\phantom{\rule{0.7em}{0ex}}1+\left(-9\right)=-8\text{*}\) |
| \(-1,9\) | \(\phantom{\rule{0.4em}{0ex}}-1+9=8\) |
| \(\phantom{\rule{1em}{0ex}}3,-3\) | \(3+\left(-3\right)=0\) |
\(\begin{array}{cccc}\text{Use}\phantom{\rule{0.2em}{0ex}}1,-9\phantom{\rule{0.2em}{0ex}}\text{as coefficients of the last terms.}\hfill & & & \hfill \phantom{\rule{3.4em}{0ex}}\left(r+s\right)\left(r-9s\right)\hfill \\ \text{Check your answer.}\hfill & & & \\ \phantom{\rule{2.5em}{0ex}}\left(r-9s\right)\left(r+s\right)\hfill & & & \\ \phantom{\rule{2.5em}{0ex}}{r}^{2}+rs-9rs-9{s}^{2}\hfill & & & \\ \phantom{\rule{2.5em}{0ex}}{r}^{2}-8rs-9{s}^{2}\phantom{\rule{0.2em}{0ex}}✓\hfill & & & \end{array}\)
Example
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.
\(\begin{array}{cccc}& & & {u}^{2}-9uv-12{v}^{2}\hfill \\ \text{Note that the first terms are}\phantom{\rule{0.2em}{0ex}}u,\phantom{\rule{0.2em}{0ex}}\text{last terms contain}\phantom{\rule{0.2em}{0ex}}v.\hfill & & & \left(u_v\right)\left(u_v\right)\hfill \end{array}\)
Find the numbers that multiply to \(-12\) and add to \(-9\).
| Factors of \(-12\) | Sum of factors |
|---|---|
| \(\phantom{\rule{2em}{0ex}}1,-12\) | \(1+\left(-12\right)=-11\) |
| \(\phantom{\rule{0.5em}{0ex}}-1,12\) | \(-1+12=11\) |
| \(\phantom{\rule{1.2em}{0ex}}2,-6\) | \(2+\left(-6\right)=-4\) |
| \(-2,6\) | \(-2+6=4\) |
| \(\phantom{\rule{1.2em}{0ex}}3,-4\) | \(3+\left(-4\right)=-1\) |
| \(-3,4\) | \(-3+4=1\) |
Note there are no factor pairs that give us \(-9\) as a sum. The trinomial is prime.
Key Concepts
- Factor trinomials of the form \({x}^{2}+bx+c\)
- Write the factors as two binomials with first terms x: \(\left(x\phantom{\rule{1em}{0ex}}\right)\left(x\phantom{\rule{1em}{0ex}}\right)\).
- Find two numbers m and n that
Multiply to c, \(m·n=c\)Add to b, \(m+n=b\)
- Use m and n as the last terms of the factors: \(\left(x+m\right)\left(x+n\right)\).
- Check by multiplying the factors.
This lesson is part of:
Factoring and Factorisation I