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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)

4x2+28x+49 4 x^{2}+28 x+49 \newlineWhich of the following is equivalent to the given expression?\newlineChoose 11 answer:\newline(A) (2x+7)2 (2 x+7)^{2} \newline(B) (2x+49)2 (2 x+49)^{2} \newline(C) (4x+7)2 (4 x+7)^{2} \newline(D) (4x+49)2 (4 x+49)^{2}

Full solution

Q. 4x2+28x+49 4 x^{2}+28 x+49 \newlineWhich of the following is equivalent to the given expression?\newlineChoose 11 answer:\newline(A) (2x+7)2 (2 x+7)^{2} \newline(B) (2x+49)2 (2 x+49)^{2} \newline(C) (4x+7)2 (4 x+7)^{2} \newline(D) (4x+49)2 (4 x+49)^{2}
  1. Recognize the structure: Recognize the structure of the given expression.\newlineThe given expression 4x2+28x+494x^2 + 28x + 49 is a quadratic expression. We will try to factor it, looking for a binomial squared that matches this expression.
  2. Factor the quadratic expression: Factor the quadratic expression.\newlineWe notice that the first term 4x24x^2 is a perfect square, as (2x)2=4x2(2x)^2 = 4x^2. The last term 4949 is also a perfect square, as 72=497^2 = 49. The middle term 28x28x is twice the product of the square roots of the first and last terms, since 2×2x×7=28x2 \times 2x \times 7 = 28x. This suggests that the expression is a perfect square trinomial.
  3. Write as a binomial squared: Write the expression as a binomial squared.\newlineSince the expression fits the pattern of a perfect square trinomial, we can write it as (2x+7)2(2x + 7)^2, which expands to 4x2+2×2x×7+724x^2 + 2 \times 2x \times 7 + 7^2, which simplifies to 4x2+28x+494x^2 + 28x + 49.
  4. Verify by expanding the binomial: Verify the answer by expanding the binomial.\newlineTo ensure that (2x+7)2(2x + 7)^2 is indeed equivalent to the given expression, we expand it: (2x+7)(2x+7)=4x2+14x+14x+49=4x2+28x+49(2x + 7)(2x + 7) = 4x^2 + 14x + 14x + 49 = 4x^2 + 28x + 49. This matches the original expression exactly.

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