Question

Find the value of c in the polynomial ${x^3} + 4{x^2} + cx$ if $x + 2$ is the factor.
a. 0
b. 1
c. 2
d. 4

Answer 1

Hint: In this question, we are using the remainder theorem of the polynomial and you get the value the constant c. The remainder theorem states that when a polynomial P(x) is divided by (x-a), then remainder is given by P (a).

Complete step-by-step answer:

Let the given cubic polynomial be $P(x) = {x^3} + 4{x^2} + cx$.

The remainder theorem states that P (a) gives the remainder of the given polynomial when P(x) is divided by the (x-a).

So, the term (x-a) is a factor of the given polynomial P(x).

It must divide the given polynomial P(x) completely, we have no remainder.

That is, P (a) should be zero.

The term (x-a) is comparing with given factor (x+2), we get
-a = 2

Multiplying both sides by -1, we get
a = -2

In this question, $P(x) = {x^3} + 4{x^2} + cx$ and a =-2. So (x+2) is a factor of the given polynomial P(x).

Therefore, P (-2) = 0
${( - 2)^3} + 4{( - 2)^2} + c( - 2) = 0$
$ - 8 + 4(4) - 2c = 0$
$ - 8 + 16 - 2c = 0$
$8 - 2c = 0$
$2c = 8$

Dividing both sides by 2, we get
c=4

Hence, the correct option of the given question is option (d).

Note: The possibility for the mistake is that you might get confused between the remainder theorem and factor theorem.

The remainder theorem states that
When a polynomial P(x) is divided by (x-a), then remainder is given by P (a).

The factor theorem states that
When P(a)=0 then (x−a) is a factor of the polynomial P(x)
When (x−a) is a factor of the polynomial P(x) then P(a)=0