Factoring x^4 + 5x^3 + 8x^2 + 8x - 8 into prime quadratics

[Math wiz ya rite 09]

The quartic polynomial x^4 + 5x^3 + 8x^2 + 8x - 8 can be factored into two prime quadratics of the form (x^2 + ax + b)(x^2 + cx + d). Find (a, b, c,d ) if a, b, c, and d are integers.

 

[galactus]

See here:

http://mathforum.org/dr.math/faq/faq.cu ... tions.html

 

[Math wiz ya rite 09]

See here:

http://mathforum.org/dr.math/faq/faq.cu ... tions.html

This link is for when the quartic polynomial is equal to zero. If I use that strategy, I would be assuming that the polynomial equals zero, which I cannot do.

 

[galactus]

Of course, it's equal to zero. That's the point in factoring. To find the roots.

Suppose it were equal to 5, then your constant term in the quartic would

be 3. A whole 'nuther' equation. :roll:

 

[stapel]

The quartic polynomial x^4 + 5x^3 + 8x^2 + 8x - 8 can be factored into two prime quadratics of the form (x^2 + ax + b)(x^2 + cx + d). Find (a, b, c,d ) if a, b, c, and d are integers.

Another method:

. . .(x² + ax + b)(x² + cx + d)

. . . . .= x⁴ + (a + b)x³ + (ac + b + d)x² + (ad + bc)x + bd

. . . . .= x⁴ + 5x³ + 8x² + 8x - 8

Equating coefficients, we get:

. . . . .a + b = 5

. . . . .ac + b + d = 8

. . . . .ad + bc = 8

. . . . .bd = -8

Solve the system for the values of a, b, c, and d.

Eliz.

 

[Math wiz ya rite 09]

Of course, it's equal to zero. That's the point in factoring. To find the roots.

Suppose it were equal to 5, then your constant term in the quartic would

be 3. A whole 'nuther' equation. :roll:

x^4 + (a+c)x^3 + (b+ac+d)x^2 + (bc+ad)x +bd

By comparison i got the system of equations:
a+c = 5
b+ ac + d = 8
bc+ad = 8
bd = -8

how do i solve this system though.

 

[stapel]

x^4 + (a+c)x^3 + (b+ac+d)x^2 + (bc+ad)x +bd

By comparison i got the system of equations...

Yes; that matches what you were given earlier.

(This is part of why it is helpful to show your work when you post. When you do that, tutors won't waste your time posting what you already know or have already done, and it won't look like you're just copying. It's better all 'round.)

how do i solve this system though.

Back-substitution would probably be the way to go. Pick one of the equations, solve for one of the variables, and plug that into another of the equations. Keep going until you arrive at the solution.

If you get stuck, please show what you have done. Thank you.

Eliz.

 

[Math wiz ya rite 09]

i got [2, 4, 3, -2] = [a,b,c,d].

i checked it and it worked!


thanks everyone