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! :)

- anonymous

- jamiebookeater

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- anonymous

Cherry, do you have the answers??

- anonymous

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

- anonymous

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

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## More answers

- anonymous

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

ash, is this right?

- ash2326

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

yeah you're right,

- anonymous

kk XD

- ash2326

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

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

- ash2326

glad to help :)

- anonymous

XD Amazing as always Ash :)

- ash2326

Thanks Yamaka :$

- anonymous

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

just a minute I'll get back to you

- anonymous

okay thanks :)

- ash2326

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

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

This is the same as we solve a quadratic by factoring

- ash2326

Cherry did you get it?

- anonymous

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

- ash2326

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

- ash2326

*counter

- anonymous

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

- ash2326

welcome cherry , glad to help:D:D

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