ICSE Class 10 Maths Chapter 10 Factor Theorem

Chapter 10

Factor Theorem

Points To Remember

Polynomial: An expression of the forms p (x) = a₀ xⁿ + a₁ xⁿ⁻¹ + a₂ xⁿ⁻² + ...... + aₙ₋₁ x + aₙ where a₀, a₁, a₂, ........ aₙ₋₁, aₙ are the real numbers and a₀ ≠ 0, is called a polynomial in x of degree n.

Value of a polynomial p (x) at x = α: The value of a polynomial p (x) at x = α is obtained by substituting x = α in the given polynomial and it is rested by p (α).

Division Algorithm for polynomials: On dividing a polynomial p (x) by a polynomial d (x), let the quotient be q (x) and the remainder be r (x), then p (x) = d (x) - q (x) + r (x) where either r (x) = 0 or deg. r (x) < deg. d (x). Here p (x) is called dividend, d (x) is divisor, q (x) is quotient and r (x) is the remainder. Note: When a polynomial p (x) is divided by (x - α), then the remainder is a constant, which can be zero or non-zero.

Remainder Theorem: If a polynomial p (x) is divided by (x - α), then the remainder is p (α).

Proof: When a polynomial p (x) is divided by (x - α), then by division algorithm, we obtain quotient q (x), and a constant remainder c such that

p (x) = (x - α) - q (x) + c ...(i)

On substituting x = α in (i), we get:

p (α) = (α - α) - q (α) + c = 0 - q (α) + c = 0 + c = c

Hence, remainder = p (α).

Results: (i) When p (x) is divided by (x + α), then the remainder = p (-α).

(ii) When p (x) is divided by (ax + b), then remainder = p \(\left(-\frac{b}{a}\right)\).

Factor Theorem: Let p (x) be a polynomial and α be the real number. Then (x - α) is a factor of p (x) if p (x) = 0

Proof: When know by remainder theorem that when p (x) is divided by (x - α), then remainder = p (α). Now, if (x - α) is a factor p (x), then remainder = 0 ⟹ p (α) = 0. Hence, (x - α) is a factor of p (x) if p (α) = 0.

Results: 1. (x + 2) is a factor of p (x) if p (-α) = 0

2. (ax + b) is a factor of p (x) if p \(\left(-\frac{b}{a}\right)\) = 0.

Exercise 10 (A)

Without actual division, find the remainder when:

Q. 1. p (x) = 3x² - 5x + 7 is divided by (x - 2).

Sol. p (x) = 3x² - 5x + 7 ...(i)

Let x - 2 = 0, then x = 2

Now, substituting the value of x in (i), we get

p (2) = 3 (2)² - 5 (2) + 7

= 3 x 4 - 5 x 2 + 7

= 12 - 10 + 7 = 19 - 10 = 9

∴ Hence remainder = 9 Ans.

Q. 2. p (x) = 2x³ - 5x² + 3x - 10 is divided by (x - 3)

Sol. p (x) = 2x³ - 5x² + 3x - 10 ...(i)

Let x - 3 = 0, then x = 3

Now, substituting the value of x in (i), we get

p (3) = 2 (3)³ - 5 (3)² + 3 x 3 - 10

= 2 x 27 - 5 x 9 + 9 - 10

= 54 - 45 + 9 - 10 = 63 - 55 = 8

Hence remainder = 8 Ans.

Q. 3. p (x) = 5x³ - 12x² + 17x - 6 is divided by (x - 1).

Sol. p (x) = 5x³ - 12x² + 17x - 6 ...(i)

Let x - 1 = 0, then x = 1

Substituting the value of x in (i), we get

p (1) = 5 (1)³ - 12 (1)² + 17 (1) - 6

= 5 x 1 - 12 x 1 + 17 x 1 - 6

= 5 - 12 + 17 - 6 = 22 - 18 = 4

Hence remainder = 4 Ans.

Q. 4. p (x) = 8x³ - 16x² + 14x - 5 is divided by (2x - 1).

Sol. p (x) = 8x³ - 16x² + 14x - 5 ...(i)

Let 2x - 1 = 0 then 2x = 1

⟹ x = \(\frac{1}{2}\)

Substituting the value of x in (i), we get

p \(\left(\frac{1}{2}\right)\) = 8 \(\left(\frac{1}{2}\right)\)³ - 16 \(\left(\frac{1}{2}\right)\)² + 14 x \(\frac{1}{2}\) - 5

= 8 x \(\frac{1}{8}\) - 16 x \(\frac{1}{4}\) + 14 x \(\frac{1}{2}\) - 5

= 1 - 4 + 7 - 5 = 8 - 9 = -1

Hence remainder = -1 Ans.

Q. 5. p (x) = 9x² - 6x + 2 is divided by (3x - 2).

Sol. p (x) = 9x² - 6x + 2 ...(i)

Let 3x - 2 = 0, then 3x = 2 ⟹ x = \(\frac{2}{3}\)

Substituting the value of x in (i), we get

p \(\left(\frac{2}{3}\right)\) = 9 \(\left(\frac{2}{3}\right)\)² - 6 x \(\frac{2}{3}\) + 2

= 9 x \(\frac{4}{9}\) - 6 x \(\frac{2}{3}\) + 2

= 4 - 4 + 2 = 6 - 4 = 2

Hence remainder = 2 Ans.

Q. 6. p (x) = x³ - 2x² - 5x + 6 is divided by x + 2.

Sol. p (x) = x³ - 2x² - 5x + 6 ...(i)

Let x + 2 = 0, then x = -2

Substituting the value of x in (i), we get

p (-2) = (-2)³ - 2 (-2)² - 5 (-2) + 6

= -8 - 2 x 4 + 10 + 6

= -8 - 8 + 10 + 6 = 16 - 16 = 0

Hence remainder = 0 Ans.

Q. 7. p (x) = 8x² - 2x - 15 is divided by (2x + 3).

Sol. p (x) = 8x² - 2x - 15 ...(i)

Let 2x + 3 = 0, then 2x = -3 ⟹ x = \(-\frac{3}{2}\)

Substituting the value of x in (i), we get

p \(\left(-\frac{3}{2}\right)\) = 8 \(\left(-\frac{3}{2}\right)\)² - 2 \(\left(-\frac{3}{2}\right)\) - 15

= 8 \(\left(\frac{9}{4}\right)\) + 3 - 15

= 18 + 3 - 15 = 21 - 15 = 6

Hence remainder = 6 Ans.

Teacher's Note

Understanding remainders helps us check our polynomial division work, just like how we verify change when shopping by checking if the total is correct.

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