# 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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