Question

(x4-x3+2)÷(x+2) polynomial

Long division
synthetic division
reminder theorem
factor theorem

(Solution)

Answer 1

Given :-

x + 2 as a factor of Polynomial.

x+2)x⁴-x³+2(x³-3x²

x⁴ + 2x³

__________________

-3x³+2

-3x³ - 6x²

_________________

-6x²+2

Hence,

Remainder= -6x²+2

Answer 2

Step-by-step explanation:

Given:

$( {x}^{4} - {x}^{3} + 2) \div (x + 2)$

To find: Solution of the division by

1) Long division

2) Synthetic division

3) Reminder theorem

4) Factor theorem

Solution:

1) Long Division:

Write both divisor and dividend polynomial ,perform division as division of decimals but here every time try to cancel the first term only .

As shown below

$x + 2 \: ) \: {x}^{4} - {x}^{3} + 2 \: ( {x}^{3} - 3 {x}^{2} + 6x - 12 \\ {x}^{4} + 2 {x}^{3} \: \: \: \: \: \: \: \: \: \: \: \\ ( - ) \: \: ( - ) \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ - - - - - - \\ - 3 {x}^{3} + 2 \\ \: \: \: \: \: \: \: \: \: \: \: - 3 {x}^{3} \: \: \: \: \: \: \: - 6 {x}^{2} \\ \: \: \: \: ( + ) \: \: \: \: \: \: \: \: \: \: \: \: (+ ) \\ - - - - - - - \\ 6 {x}^{2} + 2 \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: 6 {x}^{2} \: \: \: \: \: \: \: \: \: \: + 12x \\ \: \: \: \: \: \: \: \: \: \: \: \: ( - ) \: \: \: \: \: \: \: \: \: \: \: \: \: \: ( - ) \\ \: \: \: \: \: \: \: \: - - - - - - - - \\ - 12x + 2 \\ - 12x - 24 \\ ( - ) \: \: \: ( + ) \\ - - - - - - \\ 26 \\ - - - - - -$

Quotient polynomial is

${x}^{3} - 3 {x}^{2} + 6x - 12$

Remainder is 26

2) Synthetic division:

Write the coefficients of dividend in RHS and write the value of x from divisor in LHS.

Take 1 coefficients down,

Multiply it with -2 and place the value under coefficient of x³,add both and place down,repeat the process till end.

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: 1 \: \: \: - 1 \: \: \: \: \: 0 \: \: \: \: \: \: 0\: \: \: \: \: \:2\\ x = - 2 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: - 2 \: \: \: \: 6\: \: \: - 12 \: \: \:\:24 \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:---------------- \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:\: \: \: \: \: \: \: \: \:1 \: \: \: \: - 3 \: \: \: \: 6 \: \: \: - 12\: \: \:26$

Quotient polynomial is

${x}^{3} - 3 {x}^{2} + 6x - 12$

Remainder is 26

3) Remainder Theorem:

It states that if (x-a) is divisor of polynomial p(x) than p(a) is remainder.

On putting x=-2 in the polynomial the remainder can be calculated.

${x}^{4} - {x}^{3} +2$

put x=-2

$( { - 2)}^{4} - ( { - 2)}^{3} + 2$

$= 16 + 8 + 2$

$= 26$

On division, Remainder is 26.

4) Factor theorem: It states that if (x-a) is a factor of polynomial p(x) than p(a)=0.

Put value of x from factor

$x + 2 = 0 \\ x = - 2$

in polynomial

$p( - 2) = ( { - 2)}^{4} - ( { - 2)}^{3} + 2$

$p( - 2) = 16 + 8 + 2$

$p( - 2) = 26$

Thus, x+2 is not a factor of polynomial.

Final answer:

Quotient of division is ${x}^{3} - 3 {x}^{2} + 6x - 12$

Remainder is 26.

x+2 is not a factor of ${x}^{4} - {x}^{3} + 2$

Hope it will help you.

To learn more:

1) divide the polynomial p(x) by the polynomial g(x) and find the quotient and remainder in each of the following 1) p(x)=a...

https://brainly.in/question/40399222

2) using long division method divide the following polynomials by a binomial and check your answer p raised to power 4 + p ...

https://brainly.in/question/6311029