I need help, How do we find the values of p and q when 12x^4 + 16x^3 + p x^2 +qx -1 is divisible by 4x^2 - 1. hence, find the other factors of this expression? thank youu! :)

Question

anonymous · Answer

Cherry, do you have the answers??

anonymous · Answer

lol, nvm XD
When you divide the polynomials, we can get this far without dealing with variables:
3x^2+4x

anonymous · Answer

After this, we have (p+3)x^2+(q+4)+1 left

anonymous · Answer

Since we can only get rid of the constant negative one by adding one(in long division) we need to multiply 4x^2-1 by 1....As the polynomial is divisible by 4x^2-1, the p+3=4
q+4=0
p=1
q=-4

anonymous · Answer

ash, is this right?

ash2326 · Answer

let's find the roots
of $$4x^2-1=0$$
$$x= \pm1/2$$
now these two roots will satisy the dividend $12x^4+16x^3+px^2+qx-1$
let's first substitute x=1/2 
$$12/16+16/8+p/4+q/2-1=0$$
$$p/4+q/2=-1-12/16$$
$$p+2q=-7$$
this is our 1st equation
let's substitute x=-1/2 
$$12/16-16/8+p/4-q/2-1=0$$
$$p-2q=9$$

ash2326 · Answer

yeah you're right,

anonymous · Answer

kk XD

ash2326 · Answer

solve p+2q=-7
p-2q=9
we get p=1
and q=-4
so the function is
$$12x^4+16x^3+x^2-4x-1$$
we have to find the other two roots, one factor is 4x^2-1, let's try to create this factor in the above equation
we can write the first two term as
$$ 3x^2(4x^2-1)+ 4x(4x^2-1)+3x^2+4x$$
if you simplify the above it'll be 12x^4+16x^3
now let's write the other terms
$$ 3x^2(4x^2-1)+ 4x(4x^2-1)+3x^2+4x+x^2-4x-1$$
simplifying we get
$$ 3x^2(4x^2-1)+ 4x(4x^2-1)+4x^2-1$$
let's take out the $4x^2-1$ common
$$(3x^2+4x+1)(4x^2-1)$$
now the other factors can be found by solving
$$3x^2+4x+1$$
its factors are (3x+1)(x+1)
so 
$$12x^4+16x^3+x^2-4x-1$$=
$$(3x+1)(x+1)(4x^2-1)$$

anonymous · Answer

;o thanks! the answers are.. p=1, q=-4 and the factors are (3x+1) & (x+1)

ash2326 · Answer

glad to help :)

anonymous · Answer

XD Amazing as always Ash :)

ash2326 · Answer

Thanks Yamaka :$

anonymous · Answer

Sorry, i don't really get this part: 
we have to find the other two roots, one factor is 4x^2-1, let's try to create this factor in the above equation
we can write the first two term as
3x^2(4x^2 - 1) + 4x ( 4x^2 - 1 ) +3x^2 +4x . where do u get 3x^2 +4x at the right side of the expression?

ash2326 · Answer

just a minute I'll get back to you

anonymous · Answer

okay thanks :)

ash2326 · Answer

I wrote 12x^4+16x^3 as 
3x^2(4x^2 - 1) + 4x ( 4x^2 - 1 ) +3x^2 +4x 
if you simplify this, I'll show you, notice the terms
12x^4-3x^2+ 16x^3-4x + 3x^2+4x
             __                    _       __      _
if you combine like terms
you'll get
12x^4+16x^3

ash2326 · Answer

when i write 12x^4 as 3x^2(4x^2-1) then I've a extra term generated which is -3x^2 to counter it I added a 3x^2
$12x^4=3x^2(4x^2-1)+3x^2$
similarly when I write 16x^3 as 4x(4x^2-1) ,i've got extra term -4x , to counter it I added 4x
$$16x^3=4x(4x^2-1)+4x$$
I did this because I don't want to divide. 
If you want to find the other factors when you know one or two factor, you should try to create that factor, that factor will come out , and can then easily solve the remaining to find the other factors.

ash2326 · Answer

This is the same as we solve a quadratic by factoring

ash2326 · Answer

Cherry did you get it?

anonymous · Answer

umm not quite :/ 
when you add an extra 3x^2 or 4x, won't the equation change?

ash2326 · Answer

no because the first two term have -3x^2 and -4x , adding 3x^2 and 4x will conter them

ash2326 · Answer

*counter

anonymous · Answer

oOO! ;o i think i get what u mean now! thanks ash!! :)

ash2326 · Answer

welcome cherry , glad to help:D:D