Topics
>
Algebra 1>
Factoring>
Factor Difference of Two Squares (Harder)>
Factor Difference of Two Squares (Harder) 2Factor Difference of Two Squares (Harder) 2
Try VividMath Premium to unlock full access
Time limit: 0
Quiz summary
0 of 4 questions completed
Questions:
- 1
- 2
- 3
- 4
Information
–
You have already completed the quiz before. Hence you can not start it again.
Quiz is loading...
You must sign in or sign up to start the quiz.
You have to finish following quiz, to start this quiz:
Loading...
- 1
- 2
- 3
- 4
- Answered
- Review
-
Question 1 of 4
1. Question
Factor.`2x^5-128x^3`Hint
Help VideoCorrect
Correct!
Incorrect
Factoring the Difference of Two Squares
$$\color{#00880A}{a}^2-\color{#9a00c7}{b}^2=(\color{#00880A}{a}-\color{#9a00c7}{b})(\color{#00880A}{a}+\color{#9a00c7}{b})$$First, find the Greatest Common Factor (GCF) of the two terms.Start by listing down their factors.Factors of `2x^5`: `2``times``x``times``x``times``x``timesxtimesx`Factors of `128x^3`: `2``times64times``x``times``x``times``x`Collect the common factors and multiply them all to get the GCF.GCF `=` `2``times``x``times``x``times``x` `=` `2x^3` Next, factor by placing `2x^3` outside a bracket.Also, place the given polynomial inside the bracket with each term divided by `2x^3`, then simplify.`2x^3[(2x^5div2x^3)-(128x^3div2x^3)]` `=` `2x^3(x^2-64)` Now, express both terms inside the parenthesis as perfect squares. In other words, both terms should have `2` as their exponent.`x^2-64` `=` `x^2-8^2` `8^2=64` Finally, label the values in the expression `x^2-8^2` and substitute the values into the formula given for Factoring the Difference of Two Squares.`a=x``b=8`$$\color{#00880A}{a}^2-\color{#9a00c7}{b}^2$$ `=` $$(\color{#00880A}{a}-\color{#9a00c7}{b})(\color{#00880A}{a}+\color{#9a00c7}{b})$$ $$2x^3(\color{#00880A}{x}^2-\color{#9a00c7}{8}^2)$$ `=` $$2x^3(\color{#00880A}{x}-\color{#9a00c7}{8})(\color{#00880A}{x}+\color{#9a00c7}{8})$$ `2x^3(x-8)(x+8)` -
Question 2 of 4
2. Question
Factor.`3x^2y^2-3`Hint
Help VideoCorrect
Great Work!
Incorrect
Factoring the Difference of Two Squares
$$\color{#00880A}{a}^2-\color{#9a00c7}{b}^2=(\color{#00880A}{a}-\color{#9a00c7}{b})(\color{#00880A}{a}+\color{#9a00c7}{b})$$First, find the Greatest Common Factor (GCF) of the two terms.Start by listing down their factors.Factors of `3x^2y^2`: `3``timesxtimesxtimesytimesy`Factors of `3`: `1times``3`Both `3x^2y^2` and `3` have `3` as their factor, so it is the GCF.Next, factor by placing `3` outside a bracket.Also, place the given polynomial inside the bracket with each term divided by `3`, then simplify.`3[(3x^2y^2div3)-(3div3)]` `=` `3(x^2y^2-1)` Now, express both terms inside the parenthesis as perfect squares. In other words, both terms should have `2` as their exponent.`x^2y^2-1` `=` `(xy)^2-1` `(xy)^2=x^2y^2` `=` `(xy)^2-1^2` `1^2=1` Finally, label the values in the expression `(xy)^2-1^2` and substitute the values into the formula given for Factoring the Difference of Two Squares.`a=xy``b=1`$$\color{#00880A}{a}^2-\color{#9a00c7}{b}^2$$ `=` $$(\color{#00880A}{a}-\color{#9a00c7}{b})(\color{#00880A}{a}+\color{#9a00c7}{b})$$ $$3[(\color{#00880A}{xy})^2-\color{#9a00c7}{1}^2]$$ `=` $$3(\color{#00880A}{xy}-\color{#9a00c7}{1})(\color{#00880A}{xy}+\color{#9a00c7}{1})$$ `3(xy-1)(xy+1)` -
Question 3 of 4
3. Question
Factor.`m^4-81`Hint
Help VideoCorrect
Exceptional!
Incorrect
Factoring the Difference of Two Squares
$$\color{#00880A}{a}^2-\color{#9a00c7}{b}^2=(\color{#00880A}{a}-\color{#9a00c7}{b})(\color{#00880A}{a}+\color{#9a00c7}{b})$$First, express both terms inside the parenthesis as perfect squares. In other words, both terms should have `2` as their exponent.`m^4-81` `=` `(m^2)^2-81` `(m^2)^2=m^4` `=` `(m^2)^2-9^2` `9^2=81` Next, label the values in the expression `(m^2)^2-9^2` and substitute the values into the formula given for Factoring the Difference of Two Squares.`a=m^2``b=9`$$\color{#00880A}{a}^2-\color{#9a00c7}{b}^2$$ `=` $$(\color{#00880A}{a}+\color{#9a00c7}{b})(\color{#00880A}{a}-\color{#9a00c7}{b})$$ $$(\color{#00880A}{m^2})^2-\color{#9a00c7}{9}^2$$ `=` $$(\color{#00880A}{m^2}+\color{#9a00c7}{9})(\color{#00880A}{m^2}-\color{#9a00c7}{9})$$ Now, express both terms inside the second parenthesis as perfect squares. In other words, both terms should have `2` as their exponent.`m^2-9` `=` `m^2-3^2` `3^2=9` Finally, label the values in the expression `m^2-3^2` and substitute the values into the formula given for Factoring the Difference of Two Squares.`a=m``b=3`$$\color{#00880A}{a}^2-\color{#9a00c7}{b}^2$$ `=` $$(\color{#00880A}{a}-\color{#9a00c7}{b})(\color{#00880A}{a}+\color{#9a00c7}{b})$$ $$(m^2+9)(\color{#00880A}{m}^2-\color{#9a00c7}{3}^2)$$ `=` $$(m^2+9)(\color{#00880A}{m}-\color{#9a00c7}{3})(\color{#00880A}{m}+\color{#9a00c7}{3})$$ `(m^2+9)(m-3)(m+3)` -
Question 4 of 4
4. Question
Factor.`16x^4-1`Hint
Help VideoCorrect
Nice Job!
Incorrect
Factoring the Difference of Two Squares
$$\color{#00880A}{a}^2-\color{#9a00c7}{b}^2=(\color{#00880A}{a}-\color{#9a00c7}{b})(\color{#00880A}{a}+\color{#9a00c7}{b})$$First, express both terms inside the parenthesis as perfect squares. In other words, both terms should have `2` as their exponent.`16x^4-1` `=` `(4x^2)^2-1` `(4x^2)^2=16x^4` `=` `(4x^2)^2-1^2` `1^2=1` Next, label the values in the expression `(4x^2)^2-1^2` and substitute the values into the formula given for Factoring the Difference of Two Squares.`a=4x^2``b=1`$$\color{#00880A}{a}^2-\color{#9a00c7}{b}^2$$ `=` $$(\color{#00880A}{a}+\color{#9a00c7}{b})(\color{#00880A}{a}-\color{#9a00c7}{b})$$ $$(\color{#00880A}{4x^2})^2-\color{#9a00c7}{1}^2$$ `=` $$(\color{#00880A}{4x^2}+\color{#9a00c7}{1})(\color{#00880A}{4x^2}-\color{#9a00c7}{1})$$ Now, express both terms inside the second parenthesis as perfect squares. In other words, both terms should have `2` as their exponent.`4x^2-1` `=` `(2x)^2-1` `(2x)^2=4x^2` `=` `(2x)^2-1^2` `1^2=1` Finally, label the values in the expression `(2x)^2-1^2` and substitute the values into the formula given for Factoring the Difference of Two Squares.`a=2x``b=1`$$\color{#00880A}{a}^2-\color{#9a00c7}{b}^2$$ `=` $$(\color{#00880A}{a}-\color{#9a00c7}{b})(\color{#00880A}{a}+\color{#9a00c7}{b})$$ $$(4x^2+1)[(\color{#00880A}{2x})^2-\color{#9a00c7}{1}^2]$$ `=` $$(4x^2+1)(\color{#00880A}{2x}-\color{#9a00c7}{1})(\color{#00880A}{2x}+\color{#9a00c7}{1})$$ `(4x^2+1)(2x-1)(2x+1)`
Quizzes
- Greatest Common Factor 1
- Greatest Common Factor 2
- Factor Expressions using GCF
- Factor Expressions 1
- Factor Expressions 2
- Factor Expressions with Negative Numbers
- Factor Difference of Two Squares 1
- Factor Difference of Two Squares 2
- Factor Difference of Two Squares 3
- Factor by Grouping
- Factor Difference of Two Squares (Harder) 1
- Factor Difference of Two Squares (Harder) 2
- Factor Difference of Two Squares (Harder) 3
- Factor Quadratics 1
- Factor Quadratics 2
- Factor Quadratics 3
- Factor Quadratics with Leading Coefficient more than 1 (1)
- Factor Quadratics with Leading Coefficient more than 1 (2)
- Factor Quadratics with Leading Coefficient more than 1 (3)
- Factor Quadratics (Complex)