# Math Help - Algebra with cubics

1. ## Algebra with cubics

Ok im studying for a maths exam and there is one equation i cant do..

If a factor of P(x)=-7+ax+5x^2+15x^3+bx^4 is (x^2-1) find the values of a and b?

here is the answer i cant work out how to do it any help is appreciated thanks

thanks Tarik

2. $x^2 - 1 = (x - 1)(x + 1)$.

So that means $x - 1$ and $x + 1$ are factors.

From this, you have $P(1) = 0$ and $P(-1) = 0$ by the factor theorem.

Substitute and solve the two resulting equations simultaneously.

3. ok another similiar question popped up and i tried to apply the same logic and bam i failed again :S

if (x+3) is a factor of f(x)= -x^3+bx^2+ax-18 and g(x)=ax^2+bx-75, find values of a and b

I begun with

f(-3)= -(-3)^3 + b(-3)^2+a(-3)-18
0=9+9b-3a
therefore a =3b+3

not sure what from here

4. $x + 3$ is also a factor of $g(x)$.

So evaluate $g(-3)$, set it = 0 and get another equation in terms of $a$ and $b$.

Then solve them simultaneously for $a, b$.

5. thanks alot Prove it.. ok im going to start to become a nuisance here but i am now stuck on another question.. and btw im away so i have no teachers to help me before my exam soon

the graph of y= (a)/(x-b) + c has a vertical asymptote at x=2 and a horizontal asymptote at y=-1

a.) find values of b and c
which will be b = 2 and c=1

thus new equation is y= (a)/(x-2) + 1

This graph then udergoes following transformations
. reflection in the x axis
. dilation by a factor of 3 from the x axis
.horizontal shif of 2 units right

b.) if the intersection of the two graphs is at (m,2), find the value of m

c.)hence find the equation of the transformed graph

now im very lost here...

p.s is there any good way of doing fractions?

6. Ok, first it would really help if you'd learn some LaTeX, makes reading the questions a lot easier.

I assume that you have $y = \frac{a}{x - b} + c$ with vertical asymptote at $x = 2$ and horizontal asymptote at $y = -1$.

Your $b$ value is correct, your $c$ value is not. It should be $-1$.

This is because it is a standard vertical translation of the hyperbola $y = \frac{1}{x}$. This has horizontal asymptote at $y = 0$. By subtracting $-1$, it translates everything down by $1$ unit. So the $c$ value is $-1$.

So your equation is $y = \frac{a}{x - 2} - 1$.

I'm sure you must have studied transformations, what you have had to do to translate, dilate and reflect your graph (they will be new constants appearing somewhere in the equation). Reread your notes to familiarise yourself with where these constants go to do these particular transformations.

7. oh whoops thanks, the only part im confused with is how to work out the A variable.
And how do i do latex?