x + 4
Wiki User
∙ 2017-04-28 01:14:45Anonymous
x2
(x - 14)(x - 2)
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)
yes, (7x - 4)2
(b/2)^2= 64
(x - 14)(x - 2)
yes
It is; (2x+4)^2.
prime
True
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)
7bb - 70b + 112 = 7b2 - 70b + 112 = 7(b2 - 10b + 16) = 7(b - 2)(b - 8)
yes, (7x - 4)2
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)
(x + 2)(x - 8)
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.