Math, asked by dineshkamaraj17, 11 months ago

Can(x^2 - 1) be a remainder while dividing x^4 - 3x^2 + 5x - 9 by (x^2 + 3)? . Justify your answer with reason

Answers

Answered by amithkhan805
21

No x^2-1 cannot be a remainder of the equation

Attachments:
Answered by Swarup1998
18

(x^2 - 1) cannot be remainder here.

Step-by-step explanation:

1. Long division:

x^2 + 3 ) x^4 - 3x^2 + 5x - 9 ( x^2 - 6

x^4 + 3x^2

- -

------------------------------

- 6x^2 + 5x - 9

- 6x^2 - 18

+ +

------------------------------

5x + 9

Thus (5x + 9) is the remainder on division of (x^4 - 3x^2 + 5x - 9) by (x^2 + 3).

Hence (x^2 - 1) cannot be the remainder.

2. Using the idea of polynomial:

• The divisor is (x^2 - 3) whose degree is 2 and it contains only the x^2 term and a constant 3.

• The assumed remainder is (x^2 - 1) whose degree is 2 and it contains only the x^2 term and a constant 2.

• Now, remainder - divisor

= (x^2 - 1) - (x^2 - 3)

= x^2 - 1 - x^2 + 3

= 2

You see, on division of (x^2 - 1) by (x^2 - 3), another remainder 2 is produced.

Hence, (x^2 - 1) cannot be the remainder here.

Division problems:

Question 1. Evaluate:- The polynomials P(x) = kx³ + 3x² - 3 and Q(x) = 2x³ - 5x +k, when divided by (x-4) leave the same remainder. ...

- https://brainly.in/question/9442851

Question 2. on dividing a polynomial px by x^2 -4 quotient and remainder are found to be x and 3 respectively the polynomial px is

- https://brainly.in/question/15931162

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