Without using long division, or synthetic division, prove that expression x^2 + 5x + 6 is a factor of polynomial x^4 + 5x^3 + 2x^2 – 20x – 24.

Answer 1

x^2+5x+6 has factors of x+3 and x+2, so if their roots (x = -3, -2) are plugged into the polynomial in question, we should get 0 for both roots by the Remainder Theorem:

x^4 + 5x^3 + 2x^2 – 20x – 24
(-3)^4 + 5(-3)^3 + 2(-3)^2 - 20(-3) - 24
81 - 135 + 18 + 60 - 24
-54 + 18 + 60 - 24
-36 + 60 - 24
24 - 24
0 <-- x+3 is a factor of x^4 + 5x^3 + 2x^2 – 20x – 24

x^4 + 5x^3 + 2x^2 – 20x – 24
(-2)^4 + 5(-2)^3 + 2(-2)^2 - 20(-2) - 24
16 - 40 + 8 + 40 - 24
-24 + 8 + 40 - 24
-16 + 40 - 24
24 - 24
0 <-- x+2 is a factor of x^4 + 5x^3 + 2x^2 – 20x – 24

Thus, since both x+3 and x+2 are factors of x^4 + 5x^3 + 2x^2 – 20x – 24, then it is proved that x^2+5x+6 is a factor of x^4 + 5x^3 + 2x^2 – 20x – 24

Answer 2

The devisor factors to (x+2)(x+3)
So sub -2 and -3 into the quartic ,if you get 0 for both, you have proven it