# Advanced Factoring

You can always use the quadratic formula to find two roots of a quadratic trinomial.

But often, you can find the roots more simply by factoring.

Sometimes, you can even use factoring to find the roots of a higher-order equation, like a cubic or quartic polynomial. Below, we show some special cases and how to factor them.**Example 1:**

Factor the trinomial ${x}^{3}+7{x}^{2}+10x$.

Here, $x$ is common to all the terms and therefore can be factored out.

${x}^{3}+7{x}^{2}+10x=x\left({x}^{2}+7x+10\right)$

We need to find two numbers whose sum is $7$ and whose product is $10$ to factor ${x}^{2}+7x+10$.

The numbers are $2$ and $5$.

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

Therefore, ${x}^{3}+7{x}^{2}+10x=x\left(x+2\right)\left(x+5\right)$.

**Example 2:**

Factor the trinomial ${x}^{2}{y}^{2}-5x{y}^{2}-24{y}^{2}$.

Here, ${y}^{2}$ is common to all the terms and therefore can be factored out.

${x}^{2}{y}^{2}-5x{y}^{2}-24{y}^{2}={y}^{2}\left({x}^{2}-5x-24\right)$

We need to find two numbers whose sum is $-5$ and whose product is $-24$ to factor ${x}^{2}-5x-24$.

Among the factor pairs of $-24$, the two numbers that have a sum of $-5$ are $-8$ and $2$.

So, ${x}^{2}-5x-24=\left(x-8\right)\left(x+2\right)$.

Therefore, ${x}^{2}{y}^{2}-5x{y}^{2}-24{y}^{2}={y}^{2}\left(x-8\right)\left(x+2\right)$.

**Example 3:**

Factor, ${x}^{4}+{x}^{2}-30$.

Here, you have a polynomial of order $4$. Substitute ${x}^{2}=X$ to get an equivalent quadratic polynomial ${X}^{2}+X-30$.

We need to find two numbers whose sum is $1$ and whose product is $-30$ to factor ${X}^{2}+X-30$.

Among the factor pairs of $-30$, the two numbers that have a sum of $1$ are $-5$ and $6$.

So, ${X}^{2}+X-30=\left(X-5\right)\left(X+6\right)$.

That is, ${x}^{4}+{x}^{2}-30=\left({x}^{2}-5\right)\left({x}^{2}+6\right)$.

You can use the identity ${a}^{2}-{b}^{2}=\left(a+b\right)\left(a-b\right)$ to reduce ${x}^{2}-5$ as $\left(x+\sqrt{5}\right)\left(x-\sqrt{5}\right)$.

The binomial ${x}^{2}+6$ is irreducible; it cannot be factored over the real numbers.

Therefore, ${x}^{4}+{x}^{2}-30=\left(x+\sqrt{5}\right)\left(x-\sqrt{5}\right)\left({x}^{2}+6\right)$.

**Example 4:**

Factor the polynomial ${x}^{3}-3{x}^{2}+4x-12$.

Here, none of the above methods will work!

Group the first $2$ terms and the last $2$ terms together.

${x}^{3}-3{x}^{2}+4x-12=\left({x}^{3}-3{x}^{2}\right)+\left(4x-12\right)$

Here, ${x}^{2}$ is common in the first $2$ terms and $4$ is common in the last $2$ terms. Factor them out!

$\left({x}^{3}-3{x}^{2}\right)+\left(4x-12\right)={x}^{2}\left(x-3\right)+4\left(x-3\right)$

Now, factor out $\left(x-3\right)$.

${x}^{2}\left(x-3\right)+4\left(x-3\right)=\left(x-3\right)\left({x}^{2}+4\right)$

The binomial ${x}^{2}+4$ is irreducible; it cannot be factored over the real numbers.

Therefore, ${x}^{3}-3{x}^{2}+4x-12=\left(x-3\right)\left({x}^{2}+4\right)$.

**Example 5:**

Factor the polynomial $6{x}^{2}+7xy+2{y}^{2}$.

We need to find two numbers whose product is equal to the product of the coefficients of ${x}^{2}$- and ${y}^{2}$- terms and whose sum is equal to the coefficient of the middle term. That is, two numbers whose sum is $7$ and whose product is $6$ times $2$ or $12$.

Among the factor pairs of $12$, the two numbers that have a sum of $7$ are $4$ and $3$.

Rewrite the middle term of the trinomial using the numbers.

$6{x}^{2}+7xy+2{y}^{2}=6{x}^{2}+4xy+3xy+2{y}^{2}$

Now, we have something similar to the one in example $4$. So, group the first $2$ terms and the last $2$ terms together.

$6{x}^{2}+7xy+2{y}^{2}=\left(6{x}^{2}+4xy\right)+\left(3xy+2{y}^{2}\right)$

Here, $2x$ is common in the first $2$ terms and $y$ is common in the last $2$ terms. Factor them out!

$\left(6{x}^{2}+4xy\right)+\left(3xy+2{y}^{2}\right)=2x\left(3x+2y\right)+y\left(3x+2y\right)$

Now, use the Distributive Property.

$2x\left(3x+2y\right)+y\left(3x+2y\right)=\left(3x+2y\right)\left(2x+y\right)$

Therefore, $6{x}^{2}+7xy+2{y}^{2}=\left(3x+2y\right)\left(2x+y\right)$.

See also factoring by grouping and irreducible polynomials.