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Let’s check out your problem:

Jude factored
28x^(2) as
(14 x)(2x^(2)).
Yasmin factored
28x^(2) as
(7x)(4x).
Which of them factored
28x^(2) correctly?
Choose 1 answer:
(A) Only Jude
(B) Only Yasmin
(c) Both Jude and Yasmin
(D) Neither Jude nor Yasmin

Jude factored $28 x^{2}$ as $(14 x)\left(2 x^{2}\right)$ . $\newline$ Yasmin factored $28 x^{2}$ as $(7 x)(4 x)$ . $\newline$ Which of them factored $28 x^{2}$ correctly? $\newline$ Choose $1$ answer: $\newline$ (A) Only Jude $\newline$ (B) Only Yasmin $\newline$ (C) Both Jude and Yasmin $\newline$ (D) Neither Jude nor Yasmin

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Quantities that combine to zero: word problems

Full solution

Q. Jude factored

28 x^{2}

as

(14 x)\left(2 x^{2}\right)

.

\newline

Yasmin factored

28 x^{2}

as

(7 x)(4 x)

.

\newline

Which of them factored

28 x^{2}

correctly?

\newline

Choose

1

answer:

\newline

(A) Only Jude

\newline

(B) Only Yasmin

\newline

(C) Both Jude and Yasmin

\newline

(D) Neither Jude nor Yasmin

Examine Jude's Factorization: Let's examine Jude's factorization of $28x^{2}$ : $\newline$ Jude factored $28x^{2}$ as $(14x)(2x)$ . To check if this is correct, we multiply the factors together: $\newline$ $(14x) \times (2x) = 28x^{2}$ . $\newline$ This is the original expression, so Jude's factorization is correct.
Examine Yasmin's Factorization: Now let's examine Yasmin's factorization of $28x^{2}$ : Yasmin factored $28x^{2}$ as $(7x)(4x)$ . To check if this is correct, we multiply the factors together: $(7x) \times (4x) = 28x^{2}$ . This is also the original expression, so Yasmin's factorization is correct as well.
Verify Correct Factorizations: Since both Jude and Yasmin have factored $28x^{2}$ correctly, the answer to the question is that both of them factored it correctly.

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