Question

find the remainder when p(x)=x^3_6x^2+14x_3 is divided by g(x)=x_2 and verify the result by actual division​

Answer 1

★Answer -

» To find:

remainder when p(x) = x³ - 6x²+14x-3 is divided by g(x) = x-2 .

» Solution:

x-2 = 0, x = 2

by putting the value of x = 2 in p(x)

p(2) = (2)³-6(2)²+14(2)-3

= 8-24+28-3

= (28+8)+(-24-3)

= 36-27

= 9

${\pink{\boxed{Remainder\:when\:p(x) \:is \:divided\:by\:g(x) = 9}}}$

» Verification:

$\large{ \tt{x - 2 \: \big){x}^{3} - 6 {x}^{2} +14x - 3 \big( {x}^{2} - 4x + 6 }\\ \: \: \: \: \: \tt \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: {x}^{3} - 2 {x}^{2} \\ \: \: \: \: \: \: \: \: \: \: \: .\: \: \: \: \: \: \: \: \: \: \ \tt \: ( - )\: ( + ) \\ \tt \rule{120}{1} \\ . \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: - 4 {x}^{2} + 14x \\ . \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: - 4 {x}^{2} + 8x \\ .\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: ( + ) \: \: \: \: \: ( - ) \\ \rule{145}{1} \\. \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: 6x - 3 \\ . \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: 6x - 12 \\ . \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: ( - )( + ) \\ \rule{150}{1} \\ . \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: + 9 \\ \rule{150}1}$

★Hence, verified.

Answer 2

Answer:

• Hence, when p(x) is divided by g(x), we get, the remainder = 9

Step-by-step explanation:

★ Given,

➡ Dividend, p(x) = x³ - 6x² + 14x - 3

➡ Divisor, g(x) = x - 2

★ To find :-

➡ Remainder, r(x) when p(x) is divided by g(x).

★ Solution :-

Since, here p(x) is to be divided by g(x) and we have to find the remainder, then, we can directly apply • Remainder Theorem • for easy and efficient answer.

So, by using Remainder Theorem, we get,

g(x) = x - 2

➡ x - 2 = 0

➡ x = 2

Now by applying the value of x = 2 in p(x), we get,

p(x) = x³ - 6x² + 14x - 3

➡ p(2) = (2)³ - 6(2)² + 14(2) - 3

➡ p(2) = 8 - 6(4) + 28 - 3

➡ p(2) = 8 - 24 + 28 - 3

➡ p(2) = 5 + 4

➡ p(2) = 9

➡ r(x) = 9

• Hence, when p(x) is divided by g(x), we get, the remainder = 9

★ NOTE : THE VERIFICATION BY ACTUAL DIVISION IS GIVEN IN THE ATTACHMENT BELOW. KINDLY REFER TO THAT. ★

➡'A piece of Supplementary Counsel' :

★ Remainder Theorem = The remainder theorem states that when a polynomial, f(x), is divided by a linear polynomial , x - a, the remainder of that division will be equivalent to f(a).

★ Factor Theorem = In algebra, the factor theorem is a theorem linking factors and zeros of a polynomial. It is a special case of the polynomial remainder theorem. The factor theorem states that a polynomial has a factor if and only if.