Q. 400 - Intermediate Algebra

Question

In the following exercises, factor completely using the difference of squares pattern, if possible.

$x^2-16x+64-y^2$

Step-by-Step Solution

Answer

The Factored form of given polynomial is, 

$(x-y-8)(x+y-8)$

Step 1. Given Information

We are given a polynomial, 

$x^2-16x+64-y^2$

The formula used for factoring using the difference of squares pattern is, 

$a^2-b^2=(a+b)(a-b)$

Step 2. Factorizing the polynomial

The given polynomial can be also written as, 

$x^2-16x+64-y^2=(x{)}^2-2·x·8+(8{)}^2-y^2$

Using ${(a-b)}^2=a^2-2ab+b^2$, we get

$x^2-16x+64-y^2=(x-8{)}^2-(y{)}^2$

Using $a^2-b^2=(a+b)(a-b)$, we get

$\begin{array}{l}{x}^2-16x+64-y^2=\left(\right(x-8)-y)\left(\right(x-8)+y)\\ x^2-16x+64-y^2=(x-y-8)(x+y-8)\end{array}$

Step 3. Checking the factorization by multiplying

Multiplying the factors, we get
$$\begin{gathered} (x-y-8)(x+y-8)=x^2-16x+64-y^2 \\ (x-y)(x+y)-8(x-y)-8(x+y)+64=x^2-16x+64-y^2 \\ {x}^2-y^2-8x+8y-8x-8y+64=x^2-16x+64-y^2 \\ {x}^2-16x+64-y^2=x^2-16x+64-y^2 \\ LHS=RHS \end{gathered}$$

Hence the factorization is correct.