# I am doing polynomials right now in Algebra 2 and I don't understand how to factor polynomials or find the real number solutions of equations. Here is an example of factoring the polynomials:...

Here is an example of factoring the polynomials: 2x^4+16x^2+24.

Here is an example of finding the real number solutions of equations: x^3-5x^2-9x+45.

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### 1 Answer

There are a number of techniques used to factor polynomials. The examples you provided use three of them:

(1) Factor `2x^4+16x^2+24`

The first step I always recommend is to **factor out the largest common factor** -- this can reduce the coefficients and often reduces the degree of the polynomial.

Here we notice that 2 is a factor of all three terms:

`2x^4+16x^2+24=2(x^4+8x^2+12)`

Now consider the terms in the parantheses. Could you have factored `x^2+8x+12` ? If this factors then:

`x^2+8x+12=(x+ ? )(x+? )` Since multiplying the first terms yields `x^2`

Then the missing pieces must multiply to get 12, and add to get 8. The factors of 12 whose sum is 8 are 2 and 6, so:

`x^2+8x+12=(x+2)(x+6)` . You can check this by multiplication.

Now look at `x^4+8x^2+12` . This is the same problem -- only the variable is `x^2` ; **we say that this quartic (degree 4) is in quadratic form** -- it is quadratic in `x^2` .

Thus `(x^2)^2+8x^2+12=(x^2+2)(x^2+6)`

Another way to think of this is to substitute `w=x^2` so that `x^4+8x+12=w^2+8w+12=(w+2)(w+6)=(x^2+2)(x^2+6)`

If a polynomial is in quadratic form, you can apply techniques that you know for quadratics as well as any special forms:

`x^6+8x^3+12=(x^3)^2+8x^3+12=(x^3+2)(x^3+6)`

`x^4-9=(x^2)^2-9=(x^2+9)(x^2-9)=(x^2+9)(x+3)(x-3)` using the **difference of two squares** twice.

(2) Find the real number solutions to `f(x)=x^3-5x^2-9x+45`

The solutions are the zeros (roots) of the function:

`x^3-5x^2-9x+45=0`

Often the easiest way to solve is to factor. Here we use another technique called **factoring by grouping**; basically you consider the function two terms at a time and factor out the greatest common factor of each:

`x^3-5x^2=x^2(x-5)`

`-9x+45=-9(x-5)`

Thus we can rewrite the problem as:

`x^2(x-5)-9(x-5)=0`

We can use the distributive property to write this as a single term; recall that a(b+c)=ab+ac or ab+ac=a(b+c).

In our case, a=(x-5) and b=`x^2` while c=-9, thus:

`x^2(x-5)-9(x-5)=0`

`(x-5)(x^2-9)=0`

`(x-5)(x+3)(x-3)=0` Using the difference of two squares

Then by the zero product property either:

`x-5=0` `x+3=0` `x-3=0` So the solutions are x=5,3,or -3

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