The difference of two squares is a factoring pattern for expressions in the form \(a^2-b^2\). It factors into \(\left(a+b\right)\left(a-b\right)\). These problems practice recognizing perfect squares, factoring out a GCF first, factoring expressions with variables, and using the pattern in equations and rational expressions.
Lesson
Notes
Practice Problems
Factor fully.
\(\textbf{1)}\) \( x^2-9 \)
The answer is \( (x+3)(x-3) \)
\(\,\,\,\,\,\,x^2-9\)
\(\,\,\,\,\,\,x^2-3^2\)
\(\,\,\,\,\,\,a^2-b^2=\left(a+b\right)\left(a-b\right)\)
\(\,\,\,\,\,\,\left(x+3\right)\left(x-3\right)\)
\(\textbf{2)}\) \( x^2-1 \)
The answer is \( (x+1)(x-1) \)
\(\,\,\,\,\,\,x^2-1\)
\(\,\,\,\,\,\,x^2-1^2\)
\(\,\,\,\,\,\,a^2-b^2=\left(a+b\right)\left(a-b\right)\)
\(\,\,\,\,\,\,\left(x+1\right)\left(x-1\right)\)
\(\textbf{3)}\) \( x^2-25 \)
The answer is \( (x+5)(x-5) \)
\(\,\,\,\,\,\,x^2-25\)
\(\,\,\,\,\,\,x^2-5^2\)
\(\,\,\,\,\,\,a^2-b^2=\left(a+b\right)\left(a-b\right)\)
\(\,\,\,\,\,\,\left(x+5\right)\left(x-5\right)\)
\(\textbf{4)}\) \( x^2-49 \)
The answer is \( (x+7)(x-7) \)
\(\,\,\,\,\,\,x^2-49\)
\(\,\,\,\,\,\,x^2-7^2\)
\(\,\,\,\,\,\,a^2-b^2=\left(a+b\right)\left(a-b\right)\)
\(\,\,\,\,\,\,\left(x+7\right)\left(x-7\right)\)
\(\textbf{5)}\) \( 4x^2-49 \)
The answer is \( (2x+7)(2x-7) \)
\(\,\,\,\,\,\,4x^2-49\)
\(\,\,\,\,\,\,\left(2x\right)^2-7^2\)
\(\,\,\,\,\,\,a^2-b^2=\left(a+b\right)\left(a-b\right)\)
\(\,\,\,\,\,\,\left(2x+7\right)\left(2x-7\right)\)
\(\textbf{6)}\) \( 4x^2-16y^2 \)
The answer is \( (2x+4y)(2x-4y) \)
\(\,\,\,\,\,\,4x^2-16y^2\)
\(\,\,\,\,\,\,\left(2x\right)^2-\left(4y\right)^2\)
\(\,\,\,\,\,\,a^2-b^2=\left(a+b\right)\left(a-b\right)\)
\(\,\,\,\,\,\,\left(2x+4y\right)\left(2x-4y\right)\)
\(\textbf{7)}\) \( 16x^4-49w^2z^2 \)
The answer is \( (4x^2+7wz)(4x^2-7wz) \)
\(\,\,\,\,\,\,16x^4-49w^2z^2\)
\(\,\,\,\,\,\,\left(4x^2\right)^2-\left(7wz\right)^2\)
\(\,\,\,\,\,\,a^2-b^2=\left(a+b\right)\left(a-b\right)\)
\(\,\,\,\,\,\,\left(4x^2+7wz\right)\left(4x^2-7wz\right)\)
\(\textbf{8)}\) \( 12x^3-27xy^2 \)
The answer is \( 3x(2x+3y)(2x-3y) \)
\(\,\,\,\,\,\,12x^3-27xy^2\)
\(\,\,\,\,\,\,3x\left(4x^2-9y^2\right)\)
\(\,\,\,\,\,\,3x\left(\left(2x\right)^2-\left(3y\right)^2\right)\)
\(\,\,\,\,\,\,3x\left(2x+3y\right)\left(2x-3y\right)\)
\(\textbf{9)}\) \( x^2+9 \)
Not Factorable or Already Factored
\(\,\,\,\,\,\,x^2+9\)
\(\,\,\,\,\,\,x^2+3^2\)
\(\,\,\,\,\,\,\text{This is a sum of squares, not a difference of squares.}\)
\(\,\,\,\,\,\,\text{So it is not factorable over the real numbers.}\)
\(\textbf{10)}\) \( 4x^4-9y^2 \)
The answer is \( (2x^2+3y)(2x^2-3y) \)
\(\,\,\,\,\,\,4x^4-9y^2\)
\(\,\,\,\,\,\,\left(2x^2\right)^2-\left(3y\right)^2\)
\(\,\,\,\,\,\,a^2-b^2=\left(a+b\right)\left(a-b\right)\)
\(\,\,\,\,\,\,\left(2x^2+3y\right)\left(2x^2-3y\right)\)
\(\textbf{11)}\) \( 25x^2-36 \)
The answer is \(\left(5x+6\right)\left(5x-6\right)\)
\(\,\,\,\,\,\,25x^2-36\)
\(\,\,\,\,\,\,\left(5x\right)^2-6^2\)
\(\,\,\,\,\,\,a^2-b^2=\left(a+b\right)\left(a-b\right)\)
\(\,\,\,\,\,\,\left(5x+6\right)\left(5x-6\right)\)
\(\textbf{12)}\) \( 81x^2-y^2 \)
The answer is \(\left(9x+y\right)\left(9x-y\right)\)
\(\,\,\,\,\,\,81x^2-y^2\)
\(\,\,\,\,\,\,\left(9x\right)^2-y^2\)
\(\,\,\,\,\,\,a^2-b^2=\left(a+b\right)\left(a-b\right)\)
\(\,\,\,\,\,\,\left(9x+y\right)\left(9x-y\right)\)
\(\textbf{13)}\) \( 49a^2-64b^2 \)
The answer is \(\left(7a+8b\right)\left(7a-8b\right)\)
\(\,\,\,\,\,\,49a^2-64b^2\)
\(\,\,\,\,\,\,\left(7a\right)^2-\left(8b\right)^2\)
\(\,\,\,\,\,\,a^2-b^2=\left(a+b\right)\left(a-b\right)\)
\(\,\,\,\,\,\,\left(7a+8b\right)\left(7a-8b\right)\)
\(\textbf{14)}\) \( x^4-16 \)
The answer is \(\left(x+2\right)\left(x-2\right)\left(x^2+4\right)\)
\(\,\,\,\,\,\,x^4-16\)
\(\,\,\,\,\,\,\left(x^2\right)^2-4^2\)
\(\,\,\,\,\,\,\left(x^2+4\right)\left(x^2-4\right)\)
\(\,\,\,\,\,\,x^2-4=x^2-2^2\)
\(\,\,\,\,\,\,\left(x^2+4\right)\left(x+2\right)\left(x-2\right)\)
\(\textbf{15)}\) \( 9x^4-25 \)
The answer is \(\left(3x^2+5\right)\left(3x^2-5\right)\)
\(\,\,\,\,\,\,9x^4-25\)
\(\,\,\,\,\,\,\left(3x^2\right)^2-5^2\)
\(\,\,\,\,\,\,a^2-b^2=\left(a+b\right)\left(a-b\right)\)
\(\,\,\,\,\,\,\left(3x^2+5\right)\left(3x^2-5\right)\)
\(\textbf{16)}\) Solve for x. \(\,\, \displaystyle 4= \frac{x^2-9}{3-x} \)
The answer is \(x=-7 \)
\(\,\,\,\,\,\,4=\frac{x^2-9}{3-x}\)
\(\,\,\,\,\,\,x^2-9=\left(x+3\right)\left(x-3\right)\)
\(\,\,\,\,\,\,3-x=-\left(x-3\right)\)
\(\,\,\,\,\,\,\frac{x^2-9}{3-x}=\frac{\left(x+3\right)\left(x-3\right)}{-\left(x-3\right)}\)
\(\,\,\,\,\,\,\frac{x^2-9}{3-x}=-\left(x+3\right)\)
\(\,\,\,\,\,\,4=-\left(x+3\right)\)
\(\,\,\,\,\,\,4=-x-3\)
\(\,\,\,\,\,\,x=-7\)
\(\textbf{17)}\) Factor fully. \(\,\,16x^4-81\)
The answer is \(\left(2x+3\right)\left(2x-3\right)\left(4x^2+9\right)\)
\(\,\,\,\,\,\,16x^4-81\)
\(\,\,\,\,\,\,\left(4x^2\right)^2-9^2\)
\(\,\,\,\,\,\,\left(4x^2+9\right)\left(4x^2-9\right)\)
\(\,\,\,\,\,\,4x^2-9=\left(2x\right)^2-3^2\)
\(\,\,\,\,\,\,\left(4x^2+9\right)\left(2x+3\right)\left(2x-3\right)\)
\(\textbf{18)}\) Factor fully. \(\,\,x^8-256\)
The answer is \(\left(x-2\right)\left(x+2\right)\left(x^2+4\right)\left(x^4+16\right)\)
\(\,\,\,\,\,\,x^8-256\)
\(\,\,\,\,\,\,\left(x^4\right)^2-16^2\)
\(\,\,\,\,\,\,\left(x^4+16\right)\left(x^4-16\right)\)
\(\,\,\,\,\,\,x^4-16=\left(x^2\right)^2-4^2\)
\(\,\,\,\,\,\,x^4-16=\left(x^2+4\right)\left(x^2-4\right)\)
\(\,\,\,\,\,\,x^2-4=\left(x+2\right)\left(x-2\right)\)
\(\,\,\,\,\,\,\left(x^4+16\right)\left(x^2+4\right)\left(x+2\right)\left(x-2\right)\)
\(\textbf{19)}\) Simplify. \(\,\,\displaystyle\frac{x^2-16}{x-4}\)
The answer is \(x+4\), where \(x\ne4\).
\(\,\,\,\,\,\,\frac{x^2-16}{x-4}\)
\(\,\,\,\,\,\,x^2-16=\left(x+4\right)\left(x-4\right)\)
\(\,\,\,\,\,\,\frac{x^2-16}{x-4}=\frac{\left(x+4\right)\left(x-4\right)}{x-4}\)
\(\,\,\,\,\,\,x+4\)
\(\,\,\,\,\,\,\text{The restriction is }x\ne4.\)
\(\textbf{20)}\) Solve for x. \(\,\,x^2-64=0\)
The answers are \(x=8\) and \(x=-8\).
\(\,\,\,\,\,\,x^2-64=0\)
\(\,\,\,\,\,\,x^2-8^2=0\)
\(\,\,\,\,\,\,\left(x+8\right)\left(x-8\right)=0\)
\(\,\,\,\,\,\,x+8=0\quad\text{or}\quad x-8=0\)
\(\,\,\,\,\,\,x=-8\quad\text{or}\quad x=8\)
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