How do you solve this?!?!?!?

Question

x^2y-5xy-24y

Factor the following polynomial completely


are these two not factorable

1) x^4-81

2) 4x^2 -64

Answer 1

x^2y - 5xy - 24y
= y(x^2 - 5x - 24)
= y(x - 8)(x + 3)

x^4 - 81
= (x^2 - 9)(x^2 + 9)
= (x - 3)(x + 3)(x^2 + 9)

4x^2 - 64
= 4(x^2 - 16)
= 4(x - 4)(x + 4)

Answer 2

x^2y - 5xy - 24y
= y(x^2 - 5x - 24)
= y(x^2 - 8x + 3x - 24)
= y(x(x-8) + 3(x-8))
= y(x+3)(x-8)

I guess you cant factor your first query more than this.

How I did the above factoring ?
Here's the method ...
Example : ax^2 + bx + c
You have to split b into two components. for example d & e.
d + e = b
d*e = a*c

In your question... after taking 'y' out, you have, inside the braces,
x^2 - 5x - 24, where
a = 1 i.e. for 1*x^2
b = -5 i.e. for -5x
c = -24 i.e. for -24

now, to find d & e,
d * e = 1 * -24 = -24
d*e = -24
d+e = -5

By substituting and solving ... or just by logic ... you can see that
d = -8
e = 3

as
-8*3 = -24
-8+3 = -5

and hence you get,
x^2 - 8x + 3x - 24
= x(x-8) + 3(x-8)
= (x+3)(x-8)

This might take some time to understand if you're new to this ... but once you get it, every equation becomes really easy.

And to answer your other queries ....

x^4-81
= (x^2+9)(x^2 -9)
= (x^2 +9)(x-3)(x+3)

4x^2-64
= 4(x^2-16)
= 4(x+4)(x-4)

Hope this helped.

Answer 3

x^2y-5xy-24y = y(x^2-5x-24) = y(x-8)(x+3)

x^4-81 = (x^2+9)(x^2 -9) = (x^2 +9)(x-3)(x+3)

4x^2-64 = 4(x^2-16) = 4(x+4)(x-4)

Answer 4

y (x^2 - 5x - 24) =
y (x^2 - 5x + (5/2)^2 - (5/2)^2 - 24) =
y (( x - 5/2)^2 - (25/4 + 24)) =
y ((x - 5/2)^2 - 121/4) =
y (x - 5/2 - 11/2)(x - 5/2 + 11/2) =
y (x - 8)(x + 3)

1) (x^2 - 9) (x^2 + 9)

2) (2x - 8)(2x + 8)

Answer 5

Sure they are.

1)(x^2+9)(x^2-9) = (x^2+9)(x+3)(x-3)
2)(2x+8)(2x-8)