math math math factoring the perfect square trinomials?

Question

x^2y^2+24xy+144=
5x^2-8x-85=
15x^2-67x+38=
4x^2-28xy+49y^2=

Answer 1

(xy + 12)(xy + 12)
(x - 5)(5x + 17)
(3x - 2)(5x - 19)
(2x - 7y)(2x - 7y)

But, that isn't going to help you learn it. And, by the way, my answers are all correct. (Two are incorrect above me.)

28x^2-33x-28=(7x + 4)(4x - 7)
-20t+3t^2-7=(3t + 1)(t - 7)
now find the positive value of b so that the trinomials are factorables...
2x^2+bx-7=(x- 1)(2x + 7)
b=5

4x^2-9x+c=(4x - 5)(x - 1)
c=5
-18x^3y^2+15x^2y^2+18xy^2=(-2x^2y + 3x)(9xy + 6y)
-90p^3q+81p^2q^2+9pq^3=(30p^2q + 3pq^2)(-3p + 3q)

Sorry, I just got your message. Your welcome.

Answer 2

x^2y^2 + 24xy + 144 = (xy + 12)^2

5x^2 - 8x - 85 = (x - 5)(5x + 17)

15x^2 - 67x + 38 = (3x - 2)(5x - 19)

4x^2 - 28xy + 49y^2 = (2x - 7y)^2

The first and last problems are the only ones that are perfect square trinomials.

Answer 3

Reverse FOIL

(xy)^2 + 24xy + 144

1) square root the first term and put it in the first place of each set of parentheses:

(xy ____) (xy____)

2) Divide middle term by 2:

24xy/2 = 12xy

3) Check that coeeficient of 12xy or 12 sqaures to the last term:

12^2 =144, so it does.

4)Since number 3 worked out, write the coefficient, 12 in the last spacein each set of parentheses:

(xy + 12) (xy + 12) = (xy + 12)^2

5) Check by doing FOIL if this perfect sqaure works

(xy)(xy) + 12xy + 12xy + 144 = (xy)^2 + 24xy + 144

Answer 4

wasnt able to reply to questions because you did not specify your email ad... but anyways heres the answers.

A.(xy+12)(xy+12)
B.(5X+17)(x-5)
C.(3x-2)(3x-19)
D.2x-7)(2x-7)

Answer 5

a million) x^2 - 10x +25 x^2 -5x -5x +25 x ( x-5) -5(x -5) ( x-5) ( x-5) answer 2) 9- 12mn^2 + 4m^2n^4 9- 6mn^2 -6mn^2 + 4m^2n^4 3( 3- 2mn^2) - 2mn^2 (3 - 2mn^2) ( 3- 2mn^2) ( 3- 2mn^2) answer 3) 16p^2 + 8pr + r^2 - 4q^2 16p^2 +4pr +4pr +r^2 - 4q^2 4p ( 4p + r) + r ( 4p +r) - 4q^2 ( 4p + r) ( 4p+r) - ( 2q)^2 ( 4p + r +2q) ( 4p +r -2q) answer