Factor the polynomial completely?

Question

y3 + 5y2 - 9y - 45

Answer 1

y³ + 5y² - 9y - 45 = 0

Group Factor

(y³ + 5y²) - (9y - 45 = 0

y²(y + 5) - 9(y + 5) = 0

(y² - 9)(y + 5) = 0

- - - - - - - -s-

Answer 2

To factor an equation with a cube in it, you want to group the terms in groups of two so that when the factoring is done, you are left with the same thing inside the parenthesis. That sounds very confusing, so I will show you what I mean.
If you group the y3 and 5y2 together, you get y2 * (y+5).
If you group the -9y and -45 together, you get -9 * (y+5).
So you can rewrite the polynomial as:
y2 * (y+5) - 9 * (y+5)
From there, you can factor out y+5 to get:
(y2 - 9) * (y+5)
You can also simplify y2 - 9 to get the final answer:
(y+3)*(y-3)*(y+5)

Answer 3

Hey there!

Here's the answer.

y^3+5y^2-9y-45 -->
(y^3+5y^2)-(9y+45) -->
y^2(y+5)-9(y+5) -->
(y^2-9)(y+5)

Notice that y^2-9 is a difference of two squares. Recall that a^2-b^2 factors to (a-b)(a+b). Since (y^2-9) factors to
(y+3)(y-3), the answer is (y+3)(y-3)(y+5).

Hope it helps!

Answer 4

y^2*(y+5)-9*(y+5)=

(y^2-9)*(y+5)=

(y+3)*(y-3)*(y+5)