• Feb 21st 2013, 08:32 PM
delilah
I need help with this one: x^3 +x^2 - 14x -24
• Feb 21st 2013, 08:40 PM
jakncoke
by inspection i note that $(-2)^3 + (-2)^2 - 14 * (-2) - 24 = -8 + 4 + 28 - 24 = 0$
so (x+2) is a factor of this polynomial. Dividing $x^3 + x^2 - 14 - 24$ by (x+2) i get $(x+2)(x^2-x-12)$ using quadratic formula for the 2nd deg, i get roots x = 4, x = -3, so the full factorization is (x+2)(x-4)(x+3) = $x^3 + x^2 - 14x - 24$
• Feb 21st 2013, 08:47 PM
delilah
so lets take the degree 3 polynomial you have here. $p(x) = a_3x^3 + a_2x^2 + a_1x + a_0$ that is the general form a degree 3 polynomial, where $a_3,a_2,..$ represent some number. so if a value b was a root for p(x), then p(b) = 0. By the theorem (x-b)*(some other polynomial) = p(x).