Factoring Quadratics

Factoring Quadratics (polynomial of second degree) can be done using factoring by grouping and regrouping (actually, we already saw such example in that note).

Let's try to solve some examples.

Example 1. Solve ${{x}}^{{2}}+{7}{x}+{10}$ .

First, we must somehow rewrite ${7}{x}$ . But how?

Let's try few options:

${7}{x}={4}{x}+{3}{x}$ : ${{x}}^{{2}}+{7}{x}+{10}={{x}}^{{2}}+{4}{x}+{3}{x}+{10}$ (can't factor)
${7}{x}={8}{x}-{x}$ : ${{x}}^{{2}}+{7}{x}+{10}={{x}}^{{2}}+{8}{x}-{x}+{10}$ (can't factor)
${7}{x}={2}{x}+{5}{x}$ : ${{x}}^{{2}}+{7}{x}+{10}={{x}}^{{2}}+{2}{x}+{5}{x}+{10}={x}{\left({x}+{2}\right)}+{5}{\left({x}+{2}\right)}={\left({x}+{5}\right)}{\left({x}+{2}\right)}$ (we factored common factor three times). SUCCESS.

Answer: ${{x}}^{{2}}+{7}{x}+{10}={\left({x}+{5}\right)}{\left({x}+{2}\right)}$ .

As above example showed, there can be infinite number of ways to rewrite a term. And it can be quite hard to find correct splitting.

Moreover, such "blind guessing" is often unsuccesful.

To narrow the search, let's apply the following method.

Suppose, that following quadratic is given: ${a}{{x}}^{{2}}+{b}{x}+{c}$ .

Assume, that we know, how to rewrite term:

${a}{{x}}^{{2}}+{b}{x}+{c}={a}{{x}}^{{2}}+{b}_{{1}}{x}+{b}_{{2}}{x}+{c}=$

$={a}{x}{\left({x}+\frac{{b}_{{1}}}{{a}}\right)}+{b}_{{2}}{\left({x}+\frac{{c}}{{b}_{{2}}}\right)}$ (factor)

As can be seen, there will be common factor, when $\frac{{b}_{{1}}}{{a}}=\frac{{c}}{{b}_{{2}}}$ .

This is proportion: cross multiply to get ${b}_{{1}}{b}_{{2}}={a}{c}$ .

To factor quadratic ${\color{brown}{{{a}{{x}}^{{2}}+{b}{x}+{c}}}}$ , try to find such numbers ${b}_{{1}}$ and ${b}_{{2}}$ , that ${\color{brown}{{{b}_{{1}}+{b}_{{2}}={b}}}}$ and ${\color{brown}{{{b}_{{1}}\cdot{b}_{{2}}={a}{c}}}}$ .

This significantly narrows the search, because we only need to find factors of ${a}{c}$ , and then check, whether any two of them sum up to ${b}$ .

Example 2. Factor ${{y}}^{{2}}+{8}{y}+{15}$ , using above method.

Here ${a}={1}$ , ${b}={8}$ , ${c}={15}$ .

Thus, ${a}{c}={1}\cdot{15}={15}$ .

Now, find factors of ${15}$ . They are $-{1},{1},-{3},{3},-{5},{5}$ .

Now ,we need to check all pairs of two factors, until we find one, whose sum is ${8}$ :

$-{1}+{1}={0}\ne{8}$
$-{1}-{3}=-{4}\ne{8}$
$-{1}+{3}={2}\ne{8}$
$-{1}-{5}=-{6}\ne{8}$
$-{1}+{5}={4}\ne{8}$
${3}-{5}=-{2}\ne{8}$
${3}+{5}={8}$ (SUCCESS)

So, we should rewrite ${8}{y}$ as ${3}{y}+{5}{y}$ : ${{y}}^{{2}}+{8}{y}+{15}={{y}}^{{2}}+{3}{y}+{5}{y}+{15}={y}{\left({y}+{3}\right)}+{5}{\left({y}+{3}\right)}={\left({y}+{5}\right)}{\left({y}+{3}\right)}$ .

Answer: ${{y}}^{{2}}+{8}{y}+{15}={\left({y}+{3}\right)}{\left({y}+{5}\right)}$ .

Let's do another example.

Example 3. Factor the following: ${6}{{x}}^{{2}}-{13}{x}-{5}$ .

Here ${a}={6}$ , ${b}=-{13}$ , ${c}=-{5}$ .

Thus, ${a}{c}={6}\cdot{\left(-{5}\right)}=-{30}$ .

Now, find factors of $-{30}$ . They are $\pm{1},\pm{2},\pm{3},\pm{5},\pm{6},\pm{10},\pm{15},\pm{30}$ .

Let's try some pairs:

$-{10}+{5}=-{5}\ne-{13}$
${15}-{30}=-{15}\ne-{13}$
${2}-{15}=-{13}$ (SUCCESS)

So, we rewrite $-{13}{x}$ as $-{15}{x}+{2}{x}$ : ${6}{{x}}^{{2}}-{13}{x}-{5}={6}{{x}}^{{2}}-{15}{x}+{2}{x}-{5}={3}{x}{\left({2}{x}-{5}\right)}+{1}{\left({2}{x}-{5}\right)}={\left({3}{x}+{1}\right)}{\left({2}{x}-{5}\right)}$ .

Answer: ${6}{{x}}^{{2}}-{13}{x}-{5}={\left({3}{x}+{1}\right)}{\left({2}{x}-{5}\right)}$ .

As above examples showed, if ${a}{c}$ is very big, then there will be a lot of factors and it is still a lot of work to do.

Luckily, there is method, that doesn't require any guessing.

To factor quadratic ${a}{{x}}^{{2}}+{b}{x}+{c}$ , find roots of the quadratic equation ${a}{{x}}^{{2}}+{b}{x}+{c}={0}$ (for example, by using quadratic fomula). Let these roots are ${p}$ and ${q}$ . Then ${\color{ma\genta}{{{a}{{x}}^{{2}}+{b}{x}+{c}={a}{\left({x}-{p}\right)}{\left({x}-{q}\right)}}}}$ .

This method directly follows from Viet Theorem.

There are three cases, depending on the number of roots:

Two Roots. ${{x}}^{{2}}-{5}{x}+{4}$ : roots are ${1}$ and ${4}$ , so ${{x}}^{{2}}-{5}{x}+{4}={\left({x}-{1}\right)}{\left({x}-{4}\right)}$ .
One Root. ${{x}}^{{2}}-{6}{x}+{9}$ : one root ${3}$ (two times), so ${{x}}^{{2}}-{6}{x}+{9}={\left({x}-{3}\right)}{\left({x}-{3}\right)}={{\left({x}-{3}\right)}}^{{2}}$ .
No Roots. ${{x}}^{{2}}+{2}{x}+{21}$ : no roots, so quadratic can't be factored.

Example 4. Factor ${21}{{x}}^{{2}}+{25}{x}-{4}$ .

Since ${a}{c}={21}\cdot{\left(-{4}\right)}=-{84}$ , then there will be a lot of factors.

Therefore, we use method, based on roots of the equation.

Solve equation ${21}{{x}}^{{2}}+{25}{x}-{4}={0}$ : roots are $\frac{{1}}{{7}}$ and $-\frac{{4}}{{3}}$ .

Thus, ${21}{{x}}^{{2}}+{25}{x}-{4}={21}{\left({x}-\frac{{1}}{{7}}\right)}{\left({x}-{\left(-\frac{{4}}{{3}}\right)}\right)}={21}{\left({x}-\frac{{1}}{{7}}\right)}{\left({x}+\frac{{4}}{{3}}\right)}$ (be careful with minus sign).

We can rewrite last expression to get rid of fractions: ${21}{\left({x}-\frac{{1}}{{7}}\right)}{\left({x}+\frac{{4}}{{3}}\right)}={7}\cdot{3}{\left({x}-\frac{{1}}{{7}}\right)}{\left({x}+\frac{{4}}{{3}}\right)}=$

$={7}{\left({x}-\frac{{1}}{{7}}\right)}\cdot{3}{\left({x}+\frac{{4}}{{3}}\right)}={\left({7}{x}-{1}\right)}{\left({3}{x}+{4}\right)}$ .

Answer: ${21}{{x}}^{{2}}+{25}{x}-{4}={\left({7}{x}-{1}\right)}{\left({3}{x}+{4}\right)}$ .

Let's now turn our attention to some cases of factoring special quadratics.

Example 5. Factor the following: ${{x}}^{{4}}+{3}{{x}}^{{2}}+{2}$ .

This doesn't seem to be standard quadratic equation.

However, it is very similar to quadratic.

Since ${{x}}^{{4}}={{x}}^{{{2}\cdot{2}}}={{\left({{x}}^{{2}}\right)}}^{{2}}$ , then ${{x}}^{{4}}+{3}{{x}}^{{2}}+{2}={{\left({{x}}^{{2}}\right)}}^{{2}}+{3}{{x}}^{{2}}+{2}$ .

So, this is quadratic equation in variable ${{x}}^{{2}}$ , not ${x}$ !

Just for visual purposes, let's substitute ${u}$ instead of ${{x}}^{{2}}$ : ${{u}}^{{2}}+{3}{u}+{2}$ .

Now, perform factoring: ${{u}}^{{2}}+{3}{u}+{2}={\left({u}+{1}\right)}{\left({u}+{2}\right)}$ .

But don't forget, that ${u}={{x}}^{{2}}$ : ${{x}}^{{4}}+{3}{{x}}^{{2}}+{2}={\left({{x}}^{{2}}+{1}\right)}{\left({{x}}^{{2}}+{2}\right)}$ .

Answer: ${{x}}^{{4}}+{3}{{x}}^{{2}}+{2}={\left({{x}}^{{2}}+{1}\right)}{\left({{x}}^{{2}}+{2}\right)}$ .

Next example is very similar to above.

Example 6. Factor ${{x}}^{{\frac{{2}}{{5}}}}+{{x}}^{{\frac{{1}}{{5}}}}-{2}$ .

Again exponent of one term is twice exponent of another term. This is sign of quadratic equation.

${{x}}^{{\frac{{2}}{{5}}}}+{{x}}^{{\frac{{1}}{{5}}}}-{2}={{\left({{x}}^{{\frac{{1}}{{5}}}}\right)}}^{{2}}+{{x}}^{{\frac{{1}}{{5}}}}-{2}$ .

Let ${u}={{x}}^{{\frac{{1}}{{5}}}}$ , then above quadratic can be rewritten as ${{u}}^{{2}}+{u}-{2}$ .

Factor it: ${{u}}^{{2}}+{u}-{2}={\left({u}+{2}\right)}{\left({u}-{1}\right)}$ .

Return to ${x}$ : ${{x}}^{{\frac{{2}}{{5}}}}+{{x}}^{{\frac{{1}}{{5}}}}-{2}={\left({{x}}^{{\frac{{1}}{{5}}}}+{2}\right)}{\left({{x}}^{{\frac{{1}}{{5}}}}-{1}\right)}$ .

Answer: ${{x}}^{{\frac{{2}}{{5}}}}+{{x}}^{{\frac{{1}}{{5}}}}-{2}={\left({{x}}^{{\frac{{1}}{{5}}}}+{2}\right)}{\left({{x}}^{{\frac{{1}}{{5}}}}-{1}\right)}$ .

Last example involves two variables, but still can be factored, using quadratic formula.

Example 7. Factor the following: ${{x}}^{{2}}+{3}{x}{y}-{10}{{y}}^{{2}}$ .

We already solved such type of problems in Factoring by Grouping and Regrouping.

Let's now solve it, using discussed method.

This is expression in two variables, but if we treat ${x}$ as variable and ${y}$ as some number, then we can apply quadratic formula.

${a}{{x}}^{{2}}+{b}{x}+{c}={{x}}^{{2}}+{3}{y}{x}-{10}{{y}}^{{2}}$ .

Here ${a}={1}$ , ${b}={3}{y}$ , ${c}=-{10}{{y}}^{{2}}$ .

Find discriminant: ${D}={{b}}^{{2}}-{4}{a}{c}={{\left({3}{y}\right)}}^{{2}}-{4}\cdot{1}\cdot{\left(-{10}{{y}}^{{2}}\right)}={9}{{y}}^{{2}}+{40}{{y}}^{{2}}={49}{{y}}^{{2}}$ .

${x}_{{1}}=\frac{{-{b}-\sqrt{{{D}}}}}{{{2}{a}}}=\frac{{-{3}{y}-\sqrt{{{49}{{y}}^{{2}}}}}}{{{2}\cdot{1}}}=\frac{{-{3}{y}-{7}{y}}}{{2}}=-{5}{y}$ .

${x}_{{2}}=\frac{{-{b}+\sqrt{{{D}}}}}{{{2}{a}}}=\frac{{-{3}{y}+\sqrt{{{49}{{y}}^{{2}}}}}}{{{2}\cdot{1}}}=\frac{{-{3}{y}+{7}{y}}}{{2}}={2}{y}$ .

Thus, ${{x}}^{{2}}+{3}{x}{y}-{10}{{y}}^{{2}}={\left({x}-{\left(-{5}{y}\right)}\right)}{\left({x}-{2}{y}\right)}={\left({x}+{5}{y}\right)}{\left({x}-{2}{y}\right)}$ .

Alternatively, we can treat ${y}$ as variable and ${x}$ as some number. Answer will be the same.

Answer: ${{x}}^{{2}}+{3}{x}{y}-{10}{{y}}^{{2}}={\left({x}+{5}{y}\right)}{\left({x}-{2}{y}\right)}$ .

Now, it is time to exercise.

Exercise 1. Factor ${{x}}^{{2}}+{4}{x}-{5}$ by guessing.

Answer: ${\left({x}+{5}\right)}{\left({x}-{1}\right)}$ .

Exercise 2. Factor ${{x}}^{{2}}-{x}-{20}$ by "smart" guessing.

Answer: ${\left({x}-{5}\right)}{\left({x}+{4}\right)}$ .

Exercise 3. Factor ${24}{{y}}^{{2}}+{26}{y}-{5}$ by "smart" guessing.

Answer: ${\left({6}{y}-{1}\right)}{\left({4}{y}+{5}\right)}$ .

Exercise 4. Factor ${18}-{19}{x}-{12}{{x}}^{{2}}$ , using quadratic formula.

Answer: $-{\left({3}{x}-{2}\right)}{\left({4}{x}+{9}\right)}$ . Hint: factor out $-{1}$ first.

Exercise 5. Factor ${16}{{x}}^{{2}}+{8}{x}+{1}$ , using quadratic formula.

Answer: ${{\left({4}{x}+{1}\right)}}^{{2}}$ .

Exercise 6. Factor ${5}{{u}}^{{2}}-{4}{u}+{5}$ , using quadartic formula.

Answer: can't be factored.

Exercise 7. Factor the following: ${{x}}^{{6}}+{12}{{x}}^{{3}}+{35}$ .

Answer: ${\left({{x}}^{{3}}+{5}\right)}{\left({{x}}^{{3}}+{7}\right)}$ . Hint: ${{x}}^{{6}}={{\left({{x}}^{{3}}\right)}}^{{2}}$ .

Exercise 8. Factor the following: ${y}+\sqrt{{{y}}}-{6}$ .

Answer: ${\left(\sqrt{{{y}}}+{3}\right)}{\left(\sqrt{{{y}}}-{2}\right)}$ . Hint: ${y}={{\left(\sqrt{{{y}}}\right)}}^{{2}}$ .

Exercise 9. Factor ${6}{{y}}^{{2}}+{x}{y}-{35}{{x}}^{{2}}$ .

Answer: ${\left({2}{y}+{5}{x}\right)}{\left({3}{y}-{7}{x}\right)}$ . Hint: treat ${y}$ as variable and ${x}$ as number, then apply quadratic formula.