Question

find and equation that has x= -2, x=5, and x= -1 as solutions

Answer 1

so the solutions are
x = -2 => (x+2) = 0
x = 5 => (x-5) = 0
x = -1 => (x+1) = 0

so those were the factors of the original polynomial
or (x+2)(x-5)(x+1) = 0

to get the equation in standard form, multiply them over

Answer 2

multiply them over?

Answer 3

well, multiply them :)

Answer 4

3 roots/solutions, means 3rd degree polynomial

Answer 5

yep. im lost

Answer 6

\(\bf (x+2)(x-5)(x+1) = 0 \implies (x+2)\times(x-5)\times(x+1) = 0 \)

Answer 7

so thats the equation?

Answer 8

i can get to x^2 - 5x + 2x - 10 (x+1) = 0

Answer 9

\(\bf (x+2)(x-5)(x+1) = 0 \implies (x^2-5x+2x-10)(x+1)=0\\\quad \\
(x^2-3x-10)(x+1)=0 \implies (x^3-3x^2-10x)+(x^2-3x-10)\)

Answer 10

\(\bf (x^2-3x-10)(x+1)=0 \implies (x^3-3x^2-10x)+(x^2-3x-10)\\\quad \\
x^3-3x^2+x^2-10x-3x-10\)

Answer 11

then just add up like-terms

Answer 12

thanks!!

Answer 13

yw

Answer 14

i ended up with \[x ^{3}-2x ^{2}-13x-10=0\]