Answer:
I think it would be 5 and 5 of each solution
Answer:
x = ± 2, x = 3
Step-by-step explanation:
note that x = 2 gives
2³ - 3(2)² - 4(2) + 12 = 8 - 12 - 8 + 12 = 0
Hence x = 2 is a root and (x - 2) is a factor
x³ - 3x² - 4x + 12 ÷ (x - 2)
= (x - 2)(x² - x - 6)
= (x - 2)(x + 2)(x - 3), hence
(x - 2)(x + 2)(x - 3) = 0
equate each factor to zero and solve for x
x - 2 = 0 ⇒ x = 2
x + 2 = 0 ⇒ x = - 2
x - 3 = 0 ⇒ x = 3
Use ac method
for ax²+bx+c=0 when a≠1
multiply a and c, let's call the product r
what 2 numbesr multiply to get r and add to get b
2y²+9y+9=0
2*9=18
wat 2 numbers multiply to get 18 and add to get 9
3 and 6
split it up to that
2y²+3y+6y+9=0
group
(2y²+3y)+(6y+9)=0
factor
y(2y+3)+3(2y+3)=0
undistribute
(y+3)(2y+3)=0
set to zero
y+3=0
y=-3
2y+3=0
2y=-3
y=-3/2
y=-3 and -3/2
it's factored form is (x+3)(2x+3)
