11.)

a.) Evaluate *j*(-1) given *j*(*x*) = 2*x*^{4} - *x*^{3} - 35*x*^{2} + 16*x* + 48.

2(-1)^{4} - (-1)^{3} - 35(-1)^{2} + 16(-1) + 48 = 0

2 + 1 - 35 - 16 + 48 = 0

b.) Explain what your answer tells you about *x* + 1 as a factor.

It tells you that when the binomial *x* + 1 is divided with *j(x*) then is it divided equally with a remainder of 0 so it is a factor.

c.) Algebraically find the remaining zeros of *j*(*x*).

GAMEMASTERX40 Mar 10, 2019

#1**+3 **

Nice presentation GameMaster, I am impressed with your new style of delivery :)

You know that x+1 is a factor so if you divide j(x) by x-1 you will find at least one more factor.

I have done it to check that it works nicely but I want you to the the division for yourself.

I just used normal algebraic long division but i suppose you could use synthetic division if you know that method.

Anyway, after you do that it is relatively easy to break the new factor into 3 to get all the factors.

Once you have the factors you can get the zeros very easily.

Give it a go and if you get stuck then explain to us what your problem is

Melody Mar 10, 2019

#2**+4 **

b.) Explain what your answer tells you about x + 1 as a factor.

It tells you that when the binomial x + 1 is divided with j(x) then is it divided equally with a remainder of 0 so it is a factor.

c.) Algebraically find the remaining zeros of j(x).

Omi67 Mar 10, 2019