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4x^2+28x+c c=?
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7?
i think it's 49
Is the question what is c for this trinomial to be the square of a binomial?
Yes.
So 49?
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\(4x^2\) is the square of \(2x\) For this to be the square of a binomial, you need: \( (2x^2 + ...~)(2x^2 + ...~) = 4x^2 + 28x + c \) Compare with: \((a + b)(a + b) = a^2 + 2ab + b^2 \) The middle term is 2ab Your middle term is 28x, and you know a = 2x. 2ab = 28x, substitute 2x for a: \(2 (2x) b = 28x \) \(b = 7\) Once you have \(b = 7\), then the last term of the trinomial is \(b^2\), so it's \(7^2 = 49\). \( 4x^2 + 28x + 49 \) \(c = 49\)
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