Creating Polynomials: cubic w/ zero at x=3 (multiplicity 2), zero at x=2, constant 18

[mathdummy99]

I am having trouble with these 3 questions:

Question 1: A cubic function with a zero at x=3 (with multiplicity of 2), a zero at x=2, and a constant term of 18
My Answer: (x-3)²(x-2)+18 = (x-2)(x²-6x+9)+18 = x³-6x²+9x-2x²+12x-18+18= x³-8x²+21x
Textbook Answer: y= -x³+8x²-21x+18

Question 2: A quartic function with zeros at x= +/-square root 2, 1, -2, and y-intercept 8.
My Answer: (x+2)(x-1)(x²-2)+4 = (x+2)(x³-2x-x²+2)+4= x⁴-x³-2x²+2x+2x³-2x²-4x+4+4= x⁴+x³-4x²​-2x+8
Textbook Answer 2: y= 2x⁴+2x³-8x²-4x+8

Question 3: a quintic function with zeroes at x= +/-1, a zero at x=2 (with multiplicity of 3) and passing through point (3,24)
My answer: (x-1)(x+1)(x-2)³ I don't know what to do with the x and y coordinates
Textbook Answer 3: 3x⁵-18x⁴+33x³-6x²-36x+24

For the first 2 questions, I have no idea what I am doing wrong. Thanks for taking time to help.

 

[Dr.Peterson]

Thanks for asking the question in exactly the right way, showing the problems, your work, and why you know it's wrong.

I am having trouble with these 3 questions:

Question 1: A cubic function with a zero at x=3 (with multiplicity of 2), a zero at x=2, and a constant term of 18
My Answer: (x-3)²(x-2)+18 = (x-2)(x²-6x+9)+18 = x³-6x²+9x-2x²+12x-18+18= x³-8x²+21x
Textbook Answer: y= -x³+8x²-21x+18

The constant term can't be something you add to your product; that changes all the zeros!

Rather take your product, (x-3)²(x-2), find the constant term you get when you expand it, and then (if necessary) multiply the function by something (which won't change the zeros) to adjust it.

Question 2:

A quartic function with zeros at x= +/-square root 2, 1, -2, and y-intercept 8.
My Answer:

(x+2)(x-1)(x²-2)+4 = (x+2)(x³-2x-x²+2)+4= x⁴-x³-2x²+2x+2x³-2x²-4x+4+4= x⁴+x³-4x²​-2x+8
Textbook Answer 2: y= 2x⁴+2x³-8x²-4x+8

Same exact issue, since the constant term is the y-intercept. Follow the same advice: expand (x+2)(x-1)(x²-2), and multiply by something if necessary.

Question 3: a quintic function with zeroes at x= +/-1, a zero at x=2 (with multiplicity of 3) and passing through point (3,24)
My answer: (x-1)(x+1)(x-2)³ I don't know what to do with the x and y coordinates
Textbook Answer 3: 3x⁵-18x⁴+33x³-6x²-36x+24

For the first 2 questions, I have no idea what I am doing wrong. Thanks for taking time to help.

As in the others, any function with these zeros and degree 5 will have the form f(x) = A(x-1)(x+1)(x-2)³. Find what f(3) is, and set A as needed to make it 24.