Question

The polynomial of degree 5, P(x) has leading coefficient 1, has roots of multiplicity 2 at x=3 and x=0, and a root of multiplicity 1 at x=-5

Answer 1

do you have to find the polynomial..?

Answer 2

yes find the polynomial. I actaully figured this one out but can you help me with this. The polynomial of degree 4, P(x) has a root of multiplicity 2 at x=3 and roots of multiplicity 1 at x=0 and x=-4. It goes through the point (5,108). . how do i find a specific point?

Answer 3

F(x)=(x-3)^2(x)(x+4)

Answer 4

yep thats good but seeing you don't know anything about the coefficient of the leading term you need

\[y = a[x(x+4)(x-3)^2]\]

substitute the values from the point to find the value of the constant a

Answer 5

oh okay, that makes sense, thank you. The polynomial of degree 3, P(x), has a root of multiplicity 2 at x=4 and a root of multiplicity 1 at x=-5. The y-intercept is y= -56 .
Find a formula for P(x) . do i plug the -56 in for the y?

Answer 6

same process use the point (0, -56)

Answer 7

i see, thanks alot

Answer 8

P(x) = a(x + 5)(x-4)^2

Answer 9

Find a degree 3 polynomial that has zeros - 4, 3 and 8 and in which the coefficient of x^2 is -14. How would i set this up? (x+4)(x-3)(x-8)=f(x) but how would i use the x^2=14?

Answer 10

ok... so you need to distribute your polynomial so you have

\[P(x) = a(x^3 -7x^2 -20x +72)\]

so you need

\[-7a = 14\]

solve for a

Answer 11

where do you get the first equation? did you distribute the problem i wrote?

Answer 12

or did you somehow incorperate the coeffient of x^2?

Answer 13

I distributed the factors from the polynomial you gave

Answer 14

oops... should be + 96

Answer 15

so then its a case of equating coefficients

Answer 16

i see, so how is the coef. of x^2=14 relevant to the problem?

Answer 17

oops.. I thought it read 14x^2

Answer 18

the question says that the coefficient of x^2 is -14

so it means the term is -14x^2

and in the expanded polynomial I did x^2 has a coefficient of -7

so the value of a must be 2

Answer 19

i see, but why did you choose 7? or was the what the original equation factored out to ?

Answer 20

i see, so how is the coef. of x^2=14 relevant to the problem?

Answer 21

ok... your polynomial in factored form is

P(x) =a[(x +4)(x-3)(x-8)]

expand the factors.... and you'll have a cubic with 4 terms...

only when you have expanded..or..distributed can you use the information about the coefficient of x^2

so the 1st part is

\[P(x) =a[(x^2 + x - 12)(x - 8)]\]

I hope this helps

Answer 22

yes, i see now. thank you