Termos comuns em fatoração

Factoring: Product of a Sum and Difference

Whenever you see a polynomial in the form of $a^2 - b^2$, where $a$ and $b$ are perfect squares, its factored form is $(a + b)(a - b)$.

Examples:

$$x^2 - 4 = (x + 2)(x - 2)$$ $$x^2 - 9x^2 = (x + 3x)(x - 3x)$$ $$x^4 - 16x^4 = (x^2 + 4x^2)(x^2 - 4x^2)$$

In the case of this final expression, we can use the same rule to factor $x^2 - 4x^2$ even further:

$$= (x^2 + 4x^2)(x + 2x)(x - 2x)$$

Test Yourself!

Great job!

This pattern will always hold for products of a sum and difference. But, as always, a good way to check your work is to multiply the binomials together to see if you get your original expression.

$$\begin{align} &=(x^2+4x)(x^2-4x)\\ &=x^4-4x^3+4x^3-16x^2\\ &=x^4-16x^2\\ \end{align}$$

As we can see, once we FOIL'd the answer, we got the original expression, so we can confirm that our answer is correct!

Try the next example!

This problem got a little tricky, so let's break down what happened.

First, we noticed that the original expression, $x^{16}-9x^{2}$ was a product of a sum and difference. So, let's use our pattern to factor easily.

We can see that $a=x^8$ and $b=3x$.

Our factored form has to take the form of $(a+b)(a-b)$. So, our factored expression is: $(x^8+3x)(x^8-3x)$.

Like before, we can check our work by multiplying out the binomials.

$$\begin{align} &=(x^8+3x)(x^8-3x)\\ &=x^{16}-3x^9+3x^9-9x^2\\ &=x^{16}-9x^2\\ \end{align}$$

(Video) Factoring Sum and Difference of Cubes

by mathman1024

Here's a video by mathman1024 showing you how to factor polynomials with the sum or difference of cubes. The formulas he uses are: $$a^3+b^3=(a+b)(a^2-ab+b^2)\\ a^3-b^3=(a-b)(a^2+ab+b^2)$$ The first polynomial he looks at is $x^3-8$. So the first step is to put it into the form of something cubed $\pm$ something else cubed. Since this problem has a minus we use minus. First put down the skeleton: $$(\ \ \ )^3-(\ \ \ )^3$$ Next we find what term cubed gives us $x^3$. Obviously the answer is $x$, so we put that in the first slot. $$(x)^3-(\ \ \ )^3$$ Then we find what cubed gives us $8$. The answer is $2$ because $2*2*2=8$. $$(x)^3-(2)^3$$ Then we factor into the form given by the equation above. In this problem $a=x$ and $b=2$. So we factor this polynomial into: $$(x-2)(x^2+2x+4)$$ Note that when using these formulas the quadratic term will not be able to factored any further. In other words, we can't factor $(x^2+2x+4)$ anymore.

The second example he goes through is $343x^3+64y^3$. See if you can solve this one on your own and then watch the video to see if your answer is correct.