Remainder Theorem | Polynomials | Assessment | English | Grade 9

Let p(x) be any polynomial of degree greater than or equal to one and let a be any real number. If p(x) is divided by the linear polynomial x – a, then the remainder is ______

• 0

• p(-a)

• p(a)

• -p(a)

Answer explanation

Remainder Theorem states that, if p(x) is any polynomial of degree greater than or equal to 1 and p(x) is divided by the linear polynomial (x – a), then the remainder is p(a).

Find the remainder when 8x² - 5x + 1 is divided by x - 2

• -1

• 0

• 21

• 23

Answer explanation

Let p(x)= 8x² - 5x + 1 the zero of the polynomial x - 2 is 2 Evaluate p(2) p(2) = 8(2)² -5(2) + 1 = 32 - 10 + 1 = 23 So, by the Remainder Theorem, 23 is the remainder when 8x² - 5x + 1 is divided by x - 2 Alternatively, we can do the long division. When we divide 8x² - 5x + 1 by x - 2, we get 23 as the remainder Hence, option 4 is the correct answer.

Find the remainder when x² + 2x - 5 is divided by 2x + 8

• 3

• 1

• 0

• 2

Answer explanation

Let p(x)= x² + 2x - 5 the zero of the polynomial 2x + 8 is -4 Evaluate p(-4) p(-4) = (-4)² + 2(-4) - 5 = 16 - 8 - 5 = 3 So, by the Remainder Theorem, 3 is the remainder when x² + 2x - 5 is divided by 2x + 8. Alternatively, we can do the long division. When we divide x² + 2x - 5 by 2x + 8, we get 3 as the remainder. Hence, option 1 is the correct answer.

Find the remainder when x³ - 1 is divided by x - 1

• 1

• 0

• -1

• 2

Answer explanation

Let p(x)= x³ - 1 the zero of the polynomial x - 1 is 1 Evaluate p(1) p(1) = (1)³ - 1 = 1 - 1 = 0 So, by the Remainder Theorem, 0 is the remainder when x³ - 1 is divided by x - 1 Alternatively, we can do the long division. When we divide x³ - 1 by x - 1, we get 0 as the remainder. Hence, option 2 is the correct answer.

Check whether the polynomial q(a) = 7a³– 5a² – a – 1 is a multiple of a - 1. Click 'Yes' when completed.

• Yes

• No

Answer explanation

q(a) will be a multiple of a - 1 only if a - 1 divides q(a) completely leaving remainder zero. Now, taking a - 1 = 0 we get, a = 1 Also, q(1) = 7(1)³ – 5(1)² – 1 – 1 = 7–5 –1–1 = 0 So the remainder obtained on dividing q(a) by a - 1 is 0 So, a - 1 is a factor of the given polynomial q(a), that is q(a) is a multiple of a - 1.

The polynomial p(x) = x⁴ - ax³ - x² - ax + 4 when divided by x -2 leaves the remainder 4. Find the value of a.

• a = 1

• a = -1

• a = 1.2

• a = -1.2

Answer explanation

The given polynomial is p(x) = x⁴ - ax³ - x² - ax + 4 Divisor = x - 2 ∴ x - 2 = 0 ∴ x = 2 Thus, by remainder theorem, p(2) = (2)⁴ - a(2)³ - (2)² - a(2) + 4 4 = 16 - 8a - 4 - 2a + 4 4 = 16 - 10a 10a = 16 + 4 10a = 12 a = 1.2