4x^(2)+28 x+49 Which of the following is equivalent to the given expression? Choose 1 answer: (A) (2x+7)^(2) (B) (2x+49)^(2) (c) (4x+7)^(2) (D) (4x+49)^(2)

Recognize the structure: Recognize the structure of the given expression. The given expression 4x^2 + 28x + 49 is a quadratic expression. We will try to factor it, looking for a binomial squared that matches this expression.Factor the quadratic expression: Factor the quadratic expression. We notice that the first term 4x^2 is a perfect square, as (2x)^2 = 4x^2. The last term 49 is also a perfect square, as 7^2 = 49. The middle term 28x is twice the product of the square roots of the first and last terms, since 2 * 2x * 7 = 28x. This suggests that the expression is a perfect square trinomial.Write as a binomial squared: Write the expression as a binomial squared. Since the expression fits the pattern of a perfect square trinomial, we can write it as (2x + 7)^2, which expands to 4x^2 + 2 * 2x * 7 + 7^2, which simplifies to 4x^2 + 28x + 49.Verify by expanding the binomial: Verify the answer by expanding the binomial. To ensure that (2x + 7)^2 is indeed equivalent to the given expression, we expand it: (2x + 7)(2x + 7) = 4x^2 + 14x + 14x + 49 = 4x^2 + 28x + 49. This matches the original expression exactly.

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4x^(2)+28 x+49
Which of the following is equivalent to the given expression?
Choose 1 answer:
(A)
(2x+7)^(2)
(B)
(2x+49)^(2)
(c)
(4x+7)^(2)
(D)
(4x+49)^(2)

$4 x^{2}+28 x+49$ $\newline$ Which of the following is equivalent to the given expression? $\newline$ Choose $1$ answer: $\newline$ (A) $(2 x+7)^{2}$ $\newline$ (B) $(2 x+49)^{2}$ $\newline$ (C) $(4 x+7)^{2}$ $\newline$ (D) $(4 x+49)^{2}$

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Algebra 1

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Full solution

4 x^{2}+28 x+49

\newline

Which of the following is equivalent to the given expression?

\newline

Choose

1

answer:

\newline

(A)

(2 x+7)^{2}

\newline

(B)

(2 x+49)^{2}

\newline

(C)

(4 x+7)^{2}

\newline

(D)

(4 x+49)^{2}

Recognize the structure: Recognize the structure of the given expression. $\newline$ The given expression $4x^2 + 28x + 49$ is a quadratic expression. We will try to factor it, looking for a binomial squared that matches this expression.
Factor the quadratic expression: Factor the quadratic expression. $\newline$ We notice that the first term $4x^2$ is a perfect square, as $(2x)^2 = 4x^2$ . The last term $49$ is also a perfect square, as $7^2 = 49$ . The middle term $28x$ is twice the product of the square roots of the first and last terms, since $2 \times 2x \times 7 = 28x$ . This suggests that the expression is a perfect square trinomial.
Write as a binomial squared: Write the expression as a binomial squared. $\newline$ Since the expression fits the pattern of a perfect square trinomial, we can write it as $(2x + 7)^2$ , which expands to $4x^2 + 2 \times 2x \times 7 + 7^2$ , which simplifies to $4x^2 + 28x + 49$ .
Verify by expanding the binomial: Verify the answer by expanding the binomial. $\newline$ To ensure that $(2x + 7)^2$ is indeed equivalent to the given expression, we expand it: $(2x + 7)(2x + 7) = 4x^2 + 14x + 14x + 49 = 4x^2 + 28x + 49$ . This matches the original expression exactly.