A community for students.
Here's the question you clicked on:
 0 viewing
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! :)
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! :)

This Question is Closed

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Cherry, do you have the answers??

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0lol, nvm XD When you divide the polynomials, we can get this far without dealing with variables: 3x^2+4x

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0After this, we have (p+3)x^2+(q+4)+1 left

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Since we can only get rid of the constant negative one by adding one(in long division) we need to multiply 4x^21 by 1....As the polynomial is divisible by 4x^21, the p+3=4 q+4=0 p=1 q=4

ash2326
 4 years ago
Best ResponseYou've already chosen the best response.0let's find the roots of \[4x^21=0\] \[x= \pm1/2\] now these two roots will satisy the dividend \(12x^4+16x^3+px^2+qx1\) let's first substitute x=1/2 \[12/16+16/8+p/4+q/21=0\] \[p/4+q/2=112/16\] \[p+2q=7\] this is our 1st equation let's substitute x=1/2 \[12/1616/8+p/4q/21=0\] \[p2q=9\]

ash2326
 4 years ago
Best ResponseYou've already chosen the best response.0solve p+2q=7 p2q=9 we get p=1 and q=4 so the function is \[12x^4+16x^3+x^24x1\] we have to find the other two roots, one factor is 4x^21, let's try to create this factor in the above equation we can write the first two term as \[ 3x^2(4x^21)+ 4x(4x^21)+3x^2+4x\] if you simplify the above it'll be 12x^4+16x^3 now let's write the other terms \[ 3x^2(4x^21)+ 4x(4x^21)+3x^2+4x+x^24x1\] simplifying we get \[ 3x^2(4x^21)+ 4x(4x^21)+4x^21\] let's take out the \(4x^21\) common \[(3x^2+4x+1)(4x^21)\] now the other factors can be found by solving \[3x^2+4x+1\] its factors are (3x+1)(x+1) so \[12x^4+16x^3+x^24x1\]= \[(3x+1)(x+1)(4x^21)\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0;o thanks! the answers are.. p=1, q=4 and the factors are (3x+1) & (x+1)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0XD Amazing as always Ash :)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Sorry, i don't really get this part: we have to find the other two roots, one factor is 4x^21, 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
 4 years ago
Best ResponseYou've already chosen the best response.0just a minute I'll get back to you

ash2326
 4 years ago
Best ResponseYou've already chosen the best response.0I 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^43x^2+ 16x^34x + 3x^2+4x __ _ __ _ if you combine like terms you'll get 12x^4+16x^3

ash2326
 4 years ago
Best ResponseYou've already chosen the best response.0when i write 12x^4 as 3x^2(4x^21) then I've a extra term generated which is 3x^2 to counter it I added a 3x^2 \(12x^4=3x^2(4x^21)+3x^2\) similarly when I write 16x^3 as 4x(4x^21) ,i've got extra term 4x , to counter it I added 4x \[16x^3=4x(4x^21)+4x\] 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
 4 years ago
Best ResponseYou've already chosen the best response.0This is the same as we solve a quadratic by factoring

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0umm not quite :/ when you add an extra 3x^2 or 4x, won't the equation change?

ash2326
 4 years ago
Best ResponseYou've already chosen the best response.0no because the first two term have 3x^2 and 4x , adding 3x^2 and 4x will conter them

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0oOO! ;o i think i get what u mean now! thanks ash!! :)

ash2326
 4 years ago
Best ResponseYou've already chosen the best response.0welcome cherry , glad to help:D:D
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.