Algebra:
Factoring

1. 
\(12{x}^{3}+11{x}^{2}+2x\)
 Solution
2. 
\(60{h}^{2}+280h+45\)
 Solution
3. 
\(8{x}^{3}-125\)
 Solution
4. 
\(-3{x}^{2}+36x-108\)
 Solution
5. 
\(3{x}^{3}+21{x}^{2}+36x\)
 Solution
6. 
\(6{x}^{2}y+4xy+2ya\)
 Solution
7. 
\({x}^{3}+5x+2{x}^{2}+10\)
 Solution

Factoring - Introduction

A polynomial is an expression composed of variables and coefficients, that involves only the operations of addition, subtraction, multiplication, and non-negative integer exponents. A common form of polynomials are quadratic expressions, which follows the form:
\(a{x}^{2}+bx+c\)
. For example:
\({x}^{2}+10x+16\)
Other variations might include extra terms, higher-power exponents or negative numbers. Regardless, in many cases, it is possible to turn the polynomial into a simpler form through a process called “factoring”. Let’s try a few simple factoring problems and find out how.

The Simple Case

The simplest case is where there is no number in front of the
\({x}^{2}\)
, meaning that the cofficient is “1”. Let's use the previous expression as our example:
\({x}^{2}+10x+16\)
To factor this expression, we look for two numbers that multiply to
\(b\)
and add up to
\(c\)
. In this case, these are
\(8\)
and
\(2\)
, which let us split the second term. This gives us:
\({x}^{2}+8x+2x+16\)
Next, we split the polynomial into two sets of terms, like this:
\(({x}^{2}+8x)+(2x+16)\)
Then we factor out the common terms to give:
\(x(x+8)+2(x+8)\)
Now, we factor out the common term again, which produces:
\((x+2)(x+8)\)

More Complex Cases

What happens if we are dealing with a more complex polynomial, such as:
\(12{x}^{3}+11{x}^{2}+2x\)
Here, we have a thrid-degree polynomial with a leading cofficient that is not
\(1\)
. Let's start by taking out the greatest common factor (GCF) of all three terms, which is
\(x\)
. This gives:
\(x(12{x}^{2}+11x+2)\)
Now, we want to do the same thing as above: Split the second term. Start by multiplying the coefficient of the first term,
\(12\)
, by the constant term,
\(2\)
. This gives
\(24\)
. Then, ask the same question as above — what numbers add up to
\(11\)
and multiply to
\(24\)
? The answer is,
\(8\)
and
\(3\)
, which gives:
\(x(12{x}^{2}+8x+3x+2)\)
Now, we can split the polynomial in two and find the common term in
\(12{x}^{2} + 8x\)
, which is
\(4x\)
. This lets us remove the factor and gives:
\(x(4x(3x+2)+(3x+2))\)
Once we factor the common term
\((3x+2)\)
, we arrive at the final answer:
\(x(3x+2)(4x+1)\)

The Cymath Advantage

Our algebra factoring calculator gives you the assistance you need when you need it. Once you have mastered the basic rules, try our factoring practice problems above and see how comfortable you are with the concept. For even more help, get full access to all “how” and “why” steps by joining Cymath Plus.