Home

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 $x^{3} + 7 x^{2} + 10 x$ .

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

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

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

The numbers are $2$ and $5$ .

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

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

Example 2:

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

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

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

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

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

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

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

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

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

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

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

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

Example 4:

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

Here, none of the above methods will work!

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

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

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

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

Now, factor out $(x - 3)$ .

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

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

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

Example 5:

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

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.

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

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

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

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

$(6 x^{2} + 4 x y) + (3 x y + 2 y^{2}) = 2 x (3 x + 2 y) + y (3 x + 2 y)$

Now, use the Distributive Property.

$2 x (3 x + 2 y) + y (3 x + 2 y) = (3 x + 2 y) (2 x + y)$

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