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Answer :
Let's check both factorizations step by step.
1. For Will's factorization, he wrote
[tex]$$7x^6 = (3x^2)(4x^4).$$[/tex]
Multiply the coefficients:
[tex]$$3 \times 4 = 12.$$[/tex]
Add the exponents of [tex]$x$[/tex]:
[tex]$$2 + 4 = 6.$$[/tex]
So, the product is
[tex]$$12x^6,$$[/tex]
which is different from [tex]$7x^6$[/tex] because [tex]$12 \neq 7$[/tex].
2. For Olivia's factorization, she wrote
[tex]$$7x^6 = (7x^2)(x^3).$$[/tex]
Multiply the coefficients:
[tex]$$7 \times 1 = 7.$$[/tex]
(Here we treat the coefficient of [tex]$x^3$[/tex] as [tex]$1$[/tex].)
Add the exponents of [tex]$x$[/tex]:
[tex]$$2 + 3 = 5.$$[/tex]
So, the product is
[tex]$$7x^5,$$[/tex]
which is different from [tex]$7x^6$[/tex] because [tex]$x^5 \neq x^6$[/tex].
Since neither product equals [tex]$7x^6$[/tex], neither Will nor Olivia factored correctly.
The correct answer is:
[tex]$$\textbf{(D) Neither Will nor Olivia.}$$[/tex]
1. For Will's factorization, he wrote
[tex]$$7x^6 = (3x^2)(4x^4).$$[/tex]
Multiply the coefficients:
[tex]$$3 \times 4 = 12.$$[/tex]
Add the exponents of [tex]$x$[/tex]:
[tex]$$2 + 4 = 6.$$[/tex]
So, the product is
[tex]$$12x^6,$$[/tex]
which is different from [tex]$7x^6$[/tex] because [tex]$12 \neq 7$[/tex].
2. For Olivia's factorization, she wrote
[tex]$$7x^6 = (7x^2)(x^3).$$[/tex]
Multiply the coefficients:
[tex]$$7 \times 1 = 7.$$[/tex]
(Here we treat the coefficient of [tex]$x^3$[/tex] as [tex]$1$[/tex].)
Add the exponents of [tex]$x$[/tex]:
[tex]$$2 + 3 = 5.$$[/tex]
So, the product is
[tex]$$7x^5,$$[/tex]
which is different from [tex]$7x^6$[/tex] because [tex]$x^5 \neq x^6$[/tex].
Since neither product equals [tex]$7x^6$[/tex], neither Will nor Olivia factored correctly.
The correct answer is:
[tex]$$\textbf{(D) Neither Will nor Olivia.}$$[/tex]
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