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Cherry, do you have the answers??
lol, nvm XD When you divide the polynomials, we can get this far without dealing with variables: 3x^2+4x
After this, we have (p+3)x^2+(q+4)+1 left
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
ash, is this right?
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 \[12/16-16/8+p/4-q/2-1=0\] \[p-2q=9\]
yeah you're right,
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 \[12x^4+16x^3+x^2-4x-1\]= \[(3x+1)(x+1)(4x^2-1)\]
;o thanks! the answers are.. p=1, q=-4 and the factors are (3x+1) & (x+1)
glad to help :)
XD Amazing as always Ash :)
Thanks Yamaka :$
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?
just a minute I'll get back to you
okay thanks :)
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
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.
This is the same as we solve a quadratic by factoring
Cherry did you get it?
umm not quite :/ when you add an extra 3x^2 or 4x, won't the equation change?
no because the first two term have -3x^2 and -4x , adding 3x^2 and 4x will conter them
oOO! ;o i think i get what u mean now! thanks ash!! :)
welcome cherry , glad to help:D:D