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Q: What binomial is a factor of this trinomial x2 plus 8x plus 16?

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(x - 14)(x - 2)

yes

prime

Yes; the factored form would be (9c+4)(9c+4) or just (9c+4)2 Since the two factors are the same, the beginning trinomial 81c2+72c+16 is a perfect square trinomial

x2 - 6x - 16 = (x + 2)(x - 8)

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(x - 14)(x - 2)

yes

It is; (2x+4)^2.

True

prime

Yes; the factored form would be (9c+4)(9c+4) or just (9c+4)2 Since the two factors are the same, the beginning trinomial 81c2+72c+16 is a perfect square trinomial

x2 - 6x - 16 = (x + 2)(x - 8)

If a trinomial is a perfect square, then the discriminant will equal 0. The discriminant is equal to B^2-4AC. The variables come from the standard form of a quadratic which is Ax^2+Bx+C In this problem, A=81, B=-72, and C=16 so the discriminant is: (-72)^2-4(81)(16)=5,184-5,184=0 so this is a perfect square trinomial. To factor, notice that 81=9^2 and 16=4^2, so 81x^2=(9x)^2. We can then factor the trinomial into (9x+4)(9x-4)

7bb - 70b + 112 = 7b2 - 70b + 112 = 7(b2 - 10b + 16) = 7(b - 2)(b - 8)

(x + 2)(x - 8)

yes, (7x - 4)2

false, in order to be one 8 would have to be a square number which it is not. So the constant term in the trinomial would need to be 4, 9, 16 etc.