Question

Use 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

Hint: Equate $g(x)$ to zero to find the value of $x$ and then substitute that value of $x$ in $p(x)$. Then use the resulting value of $p(x)$ to determine if $g(x)$ is a factor. If the resulting value is $0$, then $g(x)$ is a factor of the polynomial but if it results in any other value then it is not.

Complete step-by-step answer:
For each of the questions, we will first equate $g(x)$ to $0$ to find the value of $x$. Then we will substitute that value in $p(x)$ and check if $p(x)$ comes out to be zero. If it becomes zero then $g(x)$ is a factor, otherwise it is not a factor of $p(x)$.
(i):
Putting $g(x) = 0$ , we get,
$x + 1 = 0 \Rightarrow x = - 1$
Now substituting $x = - 1$ in $p(x)$ , we get,
$
  2{x^3} + {x^2} - 2x - 1 \\
   = 2{( - 1)^3} + {( - 1)^2} - 2( - 1) - 1 \\
   = (2 \times - 1) + 1 + 2 - 1 \\
   = - 2 + 1 + 2 - 1 \\
   = 0 \\
 $
Since $p( - 1) = 0$ , $g(x) = x + 1$ is a factor of $p(x)$.
(ii):
Putting $g(x) = 0$, we get,
 $x + 2 = 0 \Rightarrow x = - 2$
Now substituting $x = - 2$ in $p(x)$ , we get,
$
  {x^3} + 3{x^2} + 3x + 1 \\
   = {( - 2)^3} + 3{( - 2)^2} + 3( - 2) + 1 \\
   = - 8 + (3 \times 4) - 6 + 1 \\
   = - 8 + 12 - 6 + 1 \\
   = 13 - 14 \\
   = - 1 \\
 $
Since $p( - 2) \ne 0$ , $g(x) = x + 2$ is not a factor of $p(x)$.
(iii):
Putting $g(x) = 0$ , we get,
$x - 3 = 0 \Rightarrow x = 3$
Now substituting $x = 3$ in $p(x)$, we get,
$
  {x^3} - 4{x^2} + x + 6 \\
   = {(3)^3} - 4{(3)^2} + 3 + 6 \\
   = 27 - (4 \times 9) + 9 \\
   = 27 - 36 + 9 \\
   = 36 - 36 \\
   = 0 \\
 $
Since $p(3) = 0$ , $g(x) = x - 3$ is a factor of $p(x)$.

Note: To check if $g(x)$ is a factor of $p(x)$, we could also divide $p(x)$ by $g(x)$ using the long division method and check if the remainder is $0$. But using Factor theorem to verify a given factor is by far the easiest technique. Simply find the value of $x$ from the given $g(x)$ and then substitute it in $p(x)$ to determine if $g(x)$ is a factor.