Question

What is a polynomial function in standard form with zeros 1, 2, -2, and -3?

Answer 1

If the zeros are 1, 2, -2, and -3, then

x = 1, x = 2, x = -2, x = -3

x-1=0, x-2=0, x+2 = 0, x+3=0

(x-1)(x-2)(x+2)(x+3) = 0

I'll let you expand and finish up

Answer 2

(x-1)(x-2)(x+2)(x+3)=0
By expanding you will get \[x ^{4}+2*x ^{3}-7*x ^{2}-8*x+12=0\]

Answer 3

how did you get that though

Answer 4

what do you get when you expand (x-1)(x-2)

Answer 5

x^2-3x+2 ?

Answer 6

how about (x+2)(x+3)

Answer 7

x^2+5x+6

Answer 8

good on both, so that means

(x-1)(x-2)(x+2)(x+3) = 0

turns into

(x^2-3x+2)(x^2+5x+6) = 0

then you expand that to get

x^2(x^2+5x+6) - 3x(x^2+5x+6) + 2(x^2+5x+6) = 0

I'll let you do the next step

Answer 9

x^4+5x^3-3x^3-15x^2-18x+2x^2+10x+12

Answer 10

x^4+5x^3+6x^2-3x^3-15x^2-18x+2x^2+10x+12

Answer 11

x^2(x^2+5x+6) - 3x(x^2+5x+6) + 2(x^2+5x+6) = 0

x^2*x^2+x^2*5x+x^2*6 - 3x*x^2- 3x*5x- 3x*6 + 2*x^2+2*5x+2*6 = 0

x^4+5x^3+6x^2 - 3x^3- 15x^2- 18x + 2x^2+10x+12 = 0

so far, so good, keep going

Answer 12

now combine like terms and you're done

Answer 13

x^4+2x^3-7x^2-8x+12

Answer 14

you nailed it

Answer 15

Yeah ! Keep going Little_Kay

Answer 16

Thank you !

Answer 17

you're welcome