Question

given that one zero is -4, find all zeros of p(x)=X^3+x^2-22x-40

a) the root of the equation are -5,-4, and -2
b) the root of the equation are 7,-4, and 3
c) the root of the equation are 10,-4 and 2
d) the root of the equation are 5, -4 and -2

Answer 1

Assuming you know long division, divide and then factor.

Answer 2

well there is a short cut

if x = -4 is a zero then (x + 4) is a factor

and you know its a cubic so

y = (x + a)(a+b)(x + 4)



and the value of 4ab the product of the constants on the binomials is -40

so therefore

4ab = -40

or ab = -10

so either a or b is positive.... but not both

match then to your answers

y=(x -5)(x + 2)(x +4)

or

y = (x + 5)(x -2)(x+4)

find the corresponding zeros from the list

Answer 3

just do synthetic long division,
(x+4) (x^2 - 3x - 10)
(x+4) (x-5)(x+2)

the other zeros are 5 and -2

Answer 4

so it D??

Answer 5

yes, it's D

Answer 6

you don't need any division if you know about the connect to the constant in the expanded form

and yes D

Answer 7

thanks alot guys!!