Question

Find the remainder, when x4 + 4x3 - 5x2 - 6x + 7 is divided by
(i) x-3
(ii) x-1
(iii) x-2
(iv) x + 2​

Answer 1

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Here is your all answers in given attachments..

Answer 2

Answer:

The remainder when $x^4 + 4x^3 - 5x^2 - 6x + 7$ is divided by

(i) x-3 is 133

(ii) x-1 is 1

(iii) x-2 is 23

(iv) x+2 is 17

Step-by-step explanation:

By remainder theorem we have,

If p(x) is a polynomial of degree $\geq 1$ and 'a' is any real number,

Then P(a) will be the remainder when p(x) is divided by x-a.

(i) Here, p(x) = $x^4 + 4x^3 - 5x^2 - 6x + 7$ and a = 3

By Remainder theorem,  when p(x) =  $x^4 + 4x^3 - 5x^2 - 6x + 7$ is divided by x - 3

Then, the remainder = p(3) = $3^4 + 4*3^3 - 5*3^2 - 6*3 + 7$

= 81 + 108 - 45 - 18 + 7

= 133

Hence, the remainder when $x^4 + 4x^3 - 5x^2 - 6x + 7$ is divided by x - 3 is 133

(ii)  Here, p(x) = $x^4 + 4x^3 - 5x^2 - 6x + 7$ and a = 1

By Remainder theorem,  when p(x) =  $x^4 + 4x^3 - 5x^2 - 6x + 7$ is divided by x - 1

Then, the remainder = p(1) = $1^4 + 4*1^3 - 5*1^2 - 6*1 + 7$

= 1 + 4 - 5 - 6 + 7

= 1

Hence, the remainder when $x^4 + 4x^3 - 5x^2 - 6x + 7$ is divided by x - 1 is 1

(iii)  Here, p(x) = $x^4 + 4x^3 - 5x^2 - 6x + 7$ and a = 2

By Remainder theorem,  when p(x) =  $x^4 + 4x^3 - 5x^2 - 6x + 7$ is divided by x - 2

Then, the remainder = p(2) = $2^4 + 4*2^3 - 5*2^2 - 6*2 + 7$

= 16 + 32 - 20 - 12 + 7

= 23

Hence, the remainder when $x^4 + 4x^3 - 5x^2 - 6x + 7$ is divided by x - 2 is 23

(iv)  Here, p(x) = $x^4 + 4x^3 - 5x^2 - 6x + 7$ and a = -2

By Remainder theorem,  when p(x) =  $x^4 + 4x^3 - 5x^2 - 6x + 7$ is divided by x + 2

Then, the remainder = p(-2) = $(-2)^4 + 4*(-2)^3 - 5*(-2)^2 - 6*(-2) + 7$

= 16 - 32 - 20 + 12 + 7

= 17

Hence, the remainder when $x^4 + 4x^3 - 5x^2 - 6x + 7$ is divided by x - 2 is 17

Hence the reminder when $x^4 + 4x^3 - 5x^2 - 6x + 7$ is divided by

(i) x-3 is 133

(ii) x-1 is 1

(iii) x-2 is 23

(iv) x+2 is 17