Answer :

Let, f(x) = x^{3} – 7x^{2} + 6x + 4

Now,

As per the question,

x – 6 = 0

x = 6

Using Remainder theorem,

We know that when f(x) is divided by (x – 6), the remainder so obtained will be f(6).

Hence,

f(6) = (6)^{3} – 7(6)^{2} + 6(6) + 4

= (216) – 7(36) + 36 + 4

= 256 -252

= 4

Therefore,

The required remainder is 4.

Rate this question :

is divided by (x-a)

RS Aggarwal & V Aggarwal - MathematicsIn each of the following, using the remainder theorem, find the remainder when *f*(*x*) is divided by *g*(*x*):

*f*(*x*) = 4*x*^{3}-12*x*^{2}+14*x*-3, *g*(*x*) = 2*x*-1

In each of the following, using the remainder theorem, find the remainder when *f*(*x*) is divided by *g*(*x*):

*f*(*x*) = 2*x*^{4}-6*x*^{3}+2*x*^{2}-*x*+2, *g*(*x*) = *x*+2

When (x^{31} + 31) is divided by (x + 1), the remainder is

is divided by (3x+2)

RS Aggarwal & V Aggarwal - MathematicsFind the remainder when p (x) = 4x^{3} + 8x^{2} – 17x + 10 is divided by (2x – 1).

Show that:

(i) 𝑥 + 3 is a factor of 69 + 11𝑥−𝑥^{2} + 𝑥^{3}.

(ii) 2𝑥−3 is a factor of 𝑥 + 2𝑥^{3} – 9𝑥^{2} + 12

When p (x) = (x^{3} + ax^{2} + 2x + a) is divided by (x + a), the remainder is

If p (x) = 2x^{3} + ax^{2} + 3x – 5 and q (x) = x^{3} + x^{2} – 4x + a leave the same remainder when divided by (x – 2), show that

If (x^{3} + mx^{2} – x + 6) has (x - 2) as a factor and leaves a remainder r, when divided by (x - 3), find the values of m and r.