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Factor Difference of Two Squares (Harder) 3Factor Difference of Two Squares (Harder) 3
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Question 1 of 4
1. Question
Factor.`5u^45`Hint
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Factorising 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 `5u^4`: `5``timesutimesutimesutimesu`Factors of `5`: `1times``5`Both `5u^4` and `5` have `5` as their factor, so it is the GCF.Next, factor by placing `5` outside a bracket.Also, place the given polynomial inside the bracket with each term divided by `5`, then simplify.`5[(5u^4div5)(5div5)]` `=` `5(u^41)` Next, express both terms inside the parenthesis as perfect squares. In other words, both terms should have `2` as their exponent.`u^41` `=` `(u^2)^21` `(u^2)^2=u^4` `=` `(u^2)^21^2` `1^2=1` Next, label the values in the expression `(u^2)^21^2` and substitute the values into the formula given for Factoring the Difference of Two Squares.`a=u^2``b=2`$$\color{#00880A}{a}^2\color{#9a00c7}{b}^2$$ `=` $$(\color{#00880A}{a}+\color{#9a00c7}{b})(\color{#00880A}{a}\color{#9a00c7}{b})$$ $$5(\color{#00880A}{u^2})^2\color{#9a00c7}{1}^2$$ `=` $$5(\color{#00880A}{u^2}+\color{#9a00c7}{1})(\color{#00880A}{u^2}\color{#9a00c7}{1})$$ Now, express both terms inside the second parenthesis as perfect squares. In other words, both terms should have `2` as their exponent.`u^21` `=` `u^21^2` `1^2=1` Finally, label the values in the expression `u^21^2` and substitute the values into the formula given for Factoring the Difference of Two Squares.`a=u``b=1`$$\color{#00880A}{a}^2\color{#9a00c7}{b}^2$$ `=` $$(\color{#00880A}{a}\color{#9a00c7}{b})(\color{#00880A}{a}+\color{#9a00c7}{b})$$ $$5(u^2+1)(\color{#00880A}{u}^2\color{#9a00c7}{1}^2)$$ `=` $$5(u^2+1)(\color{#00880A}{u}\color{#9a00c7}{1})(\color{#00880A}{u}+\color{#9a00c7}{1})$$ `5(u^2+1)(u1)(u+1)` 
Question 2 of 4
2. Question
Factor.`(x+y)^2z^2`Hint
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Factoring the Difference of Two Squares
$$\color{#00880A}{a}^2\color{#9a00c7}{b}^2=(\color{#00880A}{a}\color{#9a00c7}{b})(\color{#00880A}{a}+\color{#9a00c7}{b})$$Label the values in the expression `(x+y)^2z^2` and substitute the values into the formula given for Factoring the Difference of Two Squares.`a=x+y``b=z`$$\color{#00880A}{a}^2\color{#9a00c7}{b}^2$$ `=` $$(\color{#00880A}{a}+\color{#9a00c7}{b})(\color{#00880A}{a}\color{#9a00c7}{b})$$ $$(\color{#00880A}{x+y})^2\color{#9a00c7}{z}^2$$ `=` $$[(\color{#00880A}{x+y})+\color{#9a00c7}{z}][(\color{#00880A}{x+y})\color{#9a00c7}{z}]$$ `=` `(x+y+z)(x+yz)` `(x+y+z)(x+yz)` 
Question 3 of 4
3. Question
Factor.`y^2(mn)^2`Hint
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Factoring the Difference of Two Squares
$$\color{#00880A}{a}^2\color{#9a00c7}{b}^2=(\color{#00880A}{a}\color{#9a00c7}{b})(\color{#00880A}{a}+\color{#9a00c7}{b})$$Label the values in the expression `y^2(mn)^2` and substitute the values into the formula given for Factoring the Difference of Two Squares.`a=y``b=mn`$$\color{#00880A}{a}^2\color{#9a00c7}{b}^2$$ `=` $$(\color{#00880A}{a}+\color{#9a00c7}{b})(\color{#00880A}{a}\color{#9a00c7}{b})$$ $$\color{#00880A}{y}^2(\color{#9a00c7}{mn})^2$$ `=` $$[\color{#00880A}{y}+(\color{#9a00c7}{mn})][\color{#00880A}{y}(\color{#9a00c7}{mn})]$$ `=` `(y+mn)(ym+n)` `(y+mn)(ym+n)` 
Question 4 of 4
4. Question
factor.`4a^24(bc)^2`Hint
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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 `4a^2`: `4``timesatimesa`Factors of `4(bc)^2`: `4``times(bc)times(bc)`Both `4a^2` and `4(bc)^2` have `4` as their factor, so it is the GCF.Next, factor by placing `4` outside a bracket.Also, place the given polynomial inside the bracket with each term divided by `4`, then simplify.`4[(4a^2div4)(4(bc)^2div4)]` `=` `4(a^2(bc)^2)` Since both terms have `2` as their exponents, they are perfect squares.Factor the expression further by using the formula given for Factor the Difference of Two Squares.`a=a``b=(bc)`$$\color{#00880A}{a}^2\color{#9a00c7}{b}^2$$ `=` $$(\color{#00880A}{a}+\color{#9a00c7}{b})(\color{#00880A}{a}\color{#9a00c7}{b})$$ $$4(\color{#00880A}{a}^2\color{#9a00c7}{(bc)}^2)$$ `=` $$4(\color{#00880A}{a}+\color{#9a00c7}{(bc)})(\color{#00880A}{a}\color{#9a00c7}{(bc)})$$ `=` `4(a+bc)(ab+c)` Remove the brackets inside `4(a+bc)(ab+c)`
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