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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( x 2 +7x+10 )

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=( x+2 )( x+5 )

Therefore, x 3 +7 x 2 +10x=x( x+2 )( x+5 ) .

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 ( x 2 5x24 )

We need to find two numbers whose sum is 5 and whose product is 24 to factor x 2 5x24 .

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

So, x 2 5x24=( x8 )( x+2 ) .

Therefore, x 2 y 2 5x y 2 24 y 2 = y 2 ( x8 )( 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 +X30 .

We need to find two numbers whose sum is 1 and whose product is 30 to factor X 2 +X30 .

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

So, X 2 +X30=( X5 )( 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 )( ab ) to reduce x 2 5 as ( x+ 5 )( x 5 ) .

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

Therefore, x 4 + x 2 30=( x+ 5 )( x 5 )( x 2 +6 ) .

Example 4:

Factor the polynomial x 3 3 x 2 +4x12 .

Here, none of the above methods will work!

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

x 3 3 x 2 +4x12=( x 3 3 x 2 )+( 4x12 )

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 )+( 4x12 )= x 2 ( x3 )+4( x3 )

Now, factor out ( x3 ) .

x 2 ( x3 )+4( x3 )=( x3 )( x 2 +4 )

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

Therefore, x 3 3 x 2 +4x12=( x3 )( x 2 +4 ) .

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 =( 6 x 2 +4xy )+( 3xy+2 y 2 )

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

( 6 x 2 +4xy )+( 3xy+2 y 2 )=2x( 3x+2y )+y( 3x+2y )

Now, use the Distributive Property.

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

Therefore, 6 x 2 +7xy+2 y 2 =( 3x+2y )( 2x+y ) .

See also factoring by grouping and irreducible polynomials.