The polynomial p(x) = x⁴ - 2x³ + 3x² - ax + 3a - 7 when divided by x + 1 leaves the remainder 19. Find the values of a. Also find the remainder when p(x) is divided by x + 2.
Solution:
Given, the polynomial is p(x) = x⁴ - 2x³ + 3x² - ax + 3a - 7
p(x) leaves the remainder when divided by x + 1.
We have to find the value of a and the remainder when p(x) is divided by x + 2.
Let g(x) = x + 1
g(x) = 0
x + 1 = 0
x = -1
Put x = -1 in p(x),
p(-1) = (-1)⁴ - 2(-1)³ + 3(-1)² - a(-1) + 3a - 7
= 1 -2(-1) + 3 + a + 3a - 7
= 1 + 2 + 3 + 4a - 7
= 6 - 7 + 4a
= 4a - 1
Given, p(-1)/g(-1) = 19
So, 4a - 1 = 19
4a = 19 + 1
4a = 20
a = 20/4
a = 5
Therefore, the value of a is 5.
p(x) = x⁴ - 2x³ + 3x² - 5x + 3(5) - 7
p(x) = x⁴ - 2x³ + 3x² - 5x + 8
Let q(x) = x + 2
Now, q(x) = 0
x + 2 = 0
x = -2
Put x = -2 in p(x),
p(-2) = (-2)⁴ - 2(-2)³ + 3(-2)² - 5(-2) + 8
= 16 -2(-8) + 3(4) + 10 + 8
= 16 + 16 + 12 + 18
= 32 + 30
= 62
Therefore, the remainder when p(x) is divided by x + 2 is 62.
✦ Try This: The polynomial p(x) = 3x⁴ + x³ + x² - ax + 3a - 11 when divided by x + 1 leaves the remainder 22. Find the values of a. Also find the remainder when p(x) is divided by x - 2.
☛ Also Check: NCERT Solutions for Class 9 Maths Chapter 2
NCERT Exemplar Class 9 Maths Exercise 2.4 Problem 2
The polynomial p(x) = x⁴ - 2x³ + 3x² - ax + 3a - 7 when divided by x + 1 leaves the remainder 19. Find the values of a. Also find the remainder when p(x) is divided by x + 2
Summary:
The polynomial p(x) = x⁴ - 2x³ + 3x² - ax + 3a - 7 when divided by x + 1 leaves the remainder 19. The value of a is 5. Also the remainder when p(x) is divided by x + 2 is 62
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