factorize help

jvignacio · #1 $Old$ 11-20-2008, 01:33 AM

hey i need help factorizing this equation. I was told to first guess and find an answer for x then work from there but i cant seem to find any values of x that can make it equal 0.

$10x^3-15x^2-4x+6$

$10x^3-15x^2-4x+6=0$

mmm4444bot · #2 $Old$ 11-20-2008, 01:47 AM

Hello jvignacio:

Rather than instruct you to guess, it would have been better for them to tell you about the Rational Roots Theorem.

Are you familiar with it?

If a polynomial has integer coefficients, then any rational roots must be of the form x = p/q where p is a factor of the constant term and q is a factor of the leading coefficient.

The constant term is 6; its factors are 1, -1, 2, -2, 3, -3, 6, and -6. (These are the possible candidates for p.)

The leading coefficient is 10; its factors are 1, -1, 2, -2, 5, -5, 10, and -10. (These are the possible candidates for q.)

Now, it's just a matter of testing the possibilities until you find one that works.

Have fun. Let us know if you need more help finding rational roots using this theorem. (Hint: there is only one rational root for the polynomial that you posted.)

Cheers,

~ Mark

jvignacio · #3 $Old$ 11-20-2008, 01:56 AM

Quote:

Originally Posted by mmm4444bot $View Post$

Hello jvignacio:

Rather than instruct you to guess, it would have been better for them to tell you about the Rational Roots Theorem.

Are you familiar with it?

If a polynomial has integer coefficients, then any rational roots must be of the form x = p/q where p is a factor of the constant term and q is a factor of the leading coefficient.

The constant term is 6; its factors are 1, -1, 2, -2, 3, -3, 6, and -6. (These are the possible candidates for p.)

The leading coefficient is 10; its factors are 1, -1, 2, -2, 5, -5, 10, and -10. (These are the possible candidates for q.)

Now, it's just a matter of testing the possibilities until you find one that works.

Have fun. Let us know if you need more help finding rational roots using this theorem. (Hint: there is only one rational root for the polynomial that you posted.)

Cheers,

~ Mark

thanks for that mark. i will give it a shot and post it back! much appreciated