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