## anonymous 4 years ago 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! :)

1. anonymous

Cherry, do you have the answers??

2. anonymous

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

3. anonymous

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

4. 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

5. anonymous

ash, is this right?

6. 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 $12/16-16/8+p/4-q/2-1=0$ $p-2q=9$

7. ash2326

yeah you're right,

8. anonymous

kk XD

9. 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 $12x^4+16x^3+x^2-4x-1$= $(3x+1)(x+1)(4x^2-1)$

10. anonymous

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

11. ash2326

12. anonymous

XD Amazing as always Ash :)

13. ash2326

Thanks Yamaka :\$

14. 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?

15. ash2326

just a minute I'll get back to you

16. anonymous

okay thanks :)

17. 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

18. 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.

19. ash2326

This is the same as we solve a quadratic by factoring

20. ash2326

Cherry did you get it?

21. anonymous

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

22. ash2326

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

23. ash2326

*counter

24. anonymous

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

25. ash2326

welcome cherry , glad to help:D:D