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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 28x2 28 x^{2} as (14x)(2x2) (14 x)\left(2 x^{2}\right) .\newlineYasmin factored 28x2 28 x^{2} as (7x)(4x) (7 x)(4 x) .\newlineWhich of them factored 28x2 28 x^{2} correctly?\newlineChoose 11 answer:\newline(A) Only Jude\newline(B) Only Yasmin\newline(C) Both Jude and Yasmin\newline(D) Neither Jude nor Yasmin

Full solution

Q. Jude factored 28x2 28 x^{2} as (14x)(2x2) (14 x)\left(2 x^{2}\right) .\newlineYasmin factored 28x2 28 x^{2} as (7x)(4x) (7 x)(4 x) .\newlineWhich of them factored 28x2 28 x^{2} correctly?\newlineChoose 11 answer:\newline(A) Only Jude\newline(B) Only Yasmin\newline(C) Both Jude and Yasmin\newline(D) Neither Jude nor Yasmin
  1. Examine Jude's Factorization: Let's examine Jude's factorization of 28x228x^{2}:\newlineJude factored 28x228x^{2} as (14x)(2x)(14x)(2x). To check if this is correct, we multiply the factors together:\newline(14x)×(2x)=28x2(14x) \times (2x) = 28x^{2}.\newlineThis is the original expression, so Jude's factorization is correct.
  2. Examine Yasmin's Factorization: Now let's examine Yasmin's factorization of 28x228x^{2}: Yasmin factored 28x228x^{2} as (7x)(4x)(7x)(4x). To check if this is correct, we multiply the factors together: (7x)×(4x)=28x2(7x) \times (4x) = 28x^{2}. This is also the original expression, so Yasmin's factorization is correct as well.
  3. Verify Correct Factorizations: Since both Jude and Yasmin have factored 28x228x^{2} correctly, the answer to the question is that both of them factored it correctly.

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