Factor 28+56 t+28 w to identify the equivalent expressions. Choose 2 answers: (A) 2(56+112 t+56 w) B 28(1+2t+w) C 4(7+13 t+7w) D 7(4+8t+4w)

Identify GCF: Identify the greatest common factor (GCF) of the terms in the trinomial 28 + 56t + 28w. The terms 28, 56t, and 28w all have a common factor of 28.Factor out GCF: Factor out the GCF from each term in the trinomial. 28 + 56t + 28w = 28(1) + 28(2t) + 28(w) = 28(1 + 2t + w)Check answer choices: Check the answer choices to see which ones are equivalent to the factored expression 28(1 + 2t + w). (A) 2(56 + 112t + 56w) is not equivalent because when factored out, it would be 2(28 + 56t + 28w), which is not the same as 28(1 + 2t + w). (B) 28(1 + 2t + w) is equivalent because it matches the factored expression we found. (C) 4(7 + 13t + 7w) is not equivalent because when expanded, it would not give the original trinomial. (D) 7(4 + 8t + 4w) is equivalent because when expanded, it gives 28 + 56t + 28w, which is the original trinomial.

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Factor $28+56 t+28 w$ to identify the equivalent expressions. $\newline$ Choose $2$ answers: $\newline$ (A) $2(56+112 t+56 w)$ $\newline$ (B) $28(1+2t+w)$ $\newline$ (C) $4(7+13 t+7w)$ $\newline$ (D) $7(4+8t+4w)$

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Math Problems

Algebra 2

Write a quadratic function from its x-intercepts and another point

Full solution

Q. Factor

28+56 t+28 w

to identify the equivalent expressions.

\newline

Choose

2

answers:

\newline

(A)

2(56+112 t+56 w)

\newline

(B)

28(1+2t+w)

\newline

(C)

4(7+13 t+7w)

\newline

(D)

7(4+8t+4w)

Identify GCF: Identify the greatest common factor (GCF) of the terms in the trinomial $28 + 56t + 28w$ . The terms $28$ , $56t$ , and $28w$ all have a common factor of $28$ .
Factor out GCF: Factor out the GCF from each term in the trinomial. $28 + 56t + 28w = 28(1) + 28(2t) + 28(w) = 28(1 + 2t + w)$
Check answer choices: Check the answer choices to see which ones are equivalent to the factored expression $28(1 + 2t + w)$ . $\newline$ (A) $2(56 + 112t + 56w)$ is not equivalent because when factored out, it would be $2(28 + 56t + 28w)$ , which is not the same as $28(1 + 2t + w)$ . $\newline$ (B) $28(1 + 2t + w)$ is equivalent because it matches the factored expression we found. $\newline$ (C) $4(7 + 13t + 7w)$ is not equivalent because when expanded, it would not give the original trinomial. $\newline$ (D) $7(4 + 8t + 4w)$ is equivalent because when expanded, it gives $28 + 56t + 28w$ , which is the original trinomial.