Question

Find the zeros of the polynomial function and state the multiplicity of each f(x) = 5(x + 7)^2(x - 7)^3

Answer 1

Set \[f(x)=5(x+7)^{2}(x-7)^{3}=5(x+7)(x+7)(x-7)(x-7)(x-7)=0\]so a product of factors equaling zero implies one or more factors are zero. \[x+7=0\]\[x=-7\]for TWO of them\[x-7=0\]\[x=7\]for THREE of them. BTW \[5\neq 0\]

Answer 2

how about the multiplicity though?

Answer 3

Yep, that's what I meant by TWO and THREE ... ;-)

Answer 4

yea but where do they go?

Answer 5

nvm thanks!

Answer 6

Well, you'd just say something like x = -7 is a zero of f(x) twice repeated, and x= 7 is a zero of f(x) thrice repeated. I guess ...

Answer 7

Hey can you help me with one more problem?

Answer 8

Yup

Answer 9

Find a cubic function with the given zeros.

Sqrt 6 , -Sqrt 6, -3

Answer 10

OK, cubic means to the third power in the independent variable, say x. Begin with \[x = 6^{1/2}\]so\[x-6^{1/2}=0\]then\[x=-6^{1/2}\]so\[x+6^{1/2}=0\]finally\[x=-3\]so\[(x+3)=0\]now combine as a product : \[g(x)=(x+6^{1/2})(x-6^{1/2})(x+3)\]which could be expanded out, but it answers the question as is.

Answer 11

Note there are others that will fit the given pattern of zeroes, any multiple of g(x) say\[h(x)=5g(x)=5(x+6^{1/2})(x-6^{1/2})(x+3)\]which is what I was hinting at when I said that \[5\neq 0\]because that factor does not alter the function's zeroes.

Answer 12

but its not one of the choices I have here

Answer 13

f(x) = x3 - 3x2 - 6x - 18

f(x) = x3 + 3x2 - 6x - 18

f(x) = x3 + 3x2 + 6x - 18

f(x) = x3 + 3x2 - 6x + 18

these are my choices..

Answer 14

What have you got?

Answer 15

Ah, we have to multiply the factors out:\[g(x)=(x-6^{1/2})(x+6^{1/2})(x+3)\]\[=(x^2-6^{1/2}x+6^{1/2}x-6)(x+3)\]\[=(x^2-6)(x+3)=x^{3}+3x^{2}-6x -18\]which is the second choice.

Answer 16

are you still there?

Answer 17

Find the remainder when f(x) is divided by (x - k)

f(x) = 4x3 - 5x2 + 2x + 11; k= 3

Answer 18

these are the choices
68

136

80

24

Answer 19

Yo, examine \[p(x)=\frac{4x^3-5x^2+2x+11}{x-3}\]are you familiar with dividing one polynomial into another ?

Answer 20

no but please I need the answer

Answer 21

and yes its long division right?

Answer 22

Yes it is, but rather difficult to compactly describe here though, I'll try ...

Answer 23

ok cool

Answer 24

almost done?

Answer 25

\[\frac{4x^3-5x^{2}+2x+11}{x-3}\]\[=\frac{4x^3-12x^2+7x^{2}+2x+11}{x-3}\]\[=\frac{4x^{2}(x-3)+7x^2+2x+11}{x-3}=4x^{2}+\frac{7x^2+2x+11}{x-3}\]\[=4x^3+\frac{7x^{2}-21x+23x+11}{x-3}=4x^3+\frac{7x(x-3)+23x+11}{x-3}\]\[=4x^3+7x+\frac{23x+11}{x-3}=4x^3+7x+\frac{23x-69+69+11}{x-3}\]\[=4x^3+7x+\frac{23(x-3)+69+11}{x-3}=4x^3+7x+23+\frac{80}{x-3}\]which we cannot divide further so 80 is the remainder.