Question

Use the Factor Theorem to determine whether g(x) is a factor of p(x) in each of the following cases:
(i) p(x)=2$x^3$+$x^2$−2x−1,g(x)=x+1
(ii) p(x)=$x^3$+3$x^2$+3x+1,g(x)=x+2
(iii) p(x)=$x^3$−4$x^2$+x+6,g(x)=x−3

Answer 1

(i) p(x)=2$x^3$+$x^2$−2x−1,g(x)=x+1

Zero of (x+1) is −1.

p(−1) = 2${(-1)}^3$+${(-1)}^2$−2(−1)−1 = −2+1+2−1 = 0

Therefore, by the Factor Theorem, (x+1) is a factor of 2$x^3$+$x^2$−2x−1.

(ii) p(x)=$x^3$+3$x^2$+3x+1,g(x)=x+2

Zero of (x+2) is −2.

p(−2) = ${(-2)}^3$+3${(-2)}^2$+3(−2)+1 = −8+12−6+1 = −1 which is not equal to 0.

Therefore, by the Factor Theorem, (x+2) is not a factor of $x^3$+3$x^2$+3x+1.

(iii) p(x)=$x^3$−4$x^2$+x+6, g(x)=x−3

Zero of (x−3) is 3.

p(3) = ${\left(3\right)}^3$−4${\left(3\right)}^2$+3+6 = 27−36+9 = 0

Therefore, by the Factor Theorem, (x−3) is the factor of $x^3$−4$x^2$+x+6.