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Learn/ 11th Class Mathematics/ Unit 9 /Numericals

Unit 9: Division of Polynomials — Exercise 9 1

11th Class Mathematics · Unit 9: Division of Polynomials

1.(i) Find remainder and quotient by dividing 3x² - x + 2 by x - 1 (ii) Find remainder and quotient by dividing x³ + 12x² - 3x + 4 by x - 2 (iii) Find remainder and quotient by dividing x⁴ - 5x³ - 8x² + 13x + 12 by x - 6 (iv) Find remainder and quotient by dividing 5x⁴ - 3x³ + 2x² - 1 by x² + 4 (v) Find remainder and quotient by dividing 3x⁴ - 5x³ + 4x - 6 by x² - 3x + 5
Given
(i) Dividend 3x2 - x + 2
Divisor x - 1
Long division begin{array}{c|cc} & 3x + 2 hline x-1 & 3x2 - x + 2 & ± 3x2 mp 3x hline & 2x + 2 & ± 2x mp 2 hline & 4 end{array}
Result
Quotient 3x + 2
Remainder 4
Given
(ii) Dividend x3 + 12x2 - 3x + 4
Divisor x - 2
First step x2(x-2) = x3 - 2x2; remainder = 14x2 - 3x + 4
Second step 14x(x-2) = 14x2 - 28x; remainder = 25x + 4
Third step 25(x-2) = 25x - 50; remainder = 54
Result
Quotient x2 + 14x + 25
Remainder 54
Given
(iii) Dividend x4 - 5x3 - 8x2 + 13x + 12
Divisor x - 6
Quotient coefficients x3, x2, -2x, 1
Result
Quotient x3 + x2 - 2x + 1
Remainder 18
Given
(iv) Dividend 5x4 - 3x3 + 2x2 - 1
Divisor x2 + 4
First quotient term 5x2(x2+4) = 5x4 + 20x2
Second quotient term -3x(x2+4) = -3x3 - 12x
Third quotient term -18(x2+4) = -18x2 - 72
Result
Quotient 5x2 - 3x - 18
Remainder 12x + 71
Given
(v) Dividend 3x4 - 5x3 + 4x - 6
Divisor x2 - 3x + 5
First quotient term 3x2(x2-3x+5) = 3x4 - 9x3 + 15x2
Second quotient term 4x(x2-3x+5) = 4x3 - 12x2 + 20x
Third quotient term -3(x2-3x+5) = -3x2 + 9x - 15
Result
Quotient 3x2 + 4x - 3
Remainder -25x + 9
2.(i) Use remainder theorem to find remainder when x² + 5x + 6 is divided by x - 2 (ii) Use remainder theorem to find remainder when x³ + 5x² + 6 is divided by x + 1 (iii) Use remainder theorem to find remainder when x⁴ + x³ + x² + x + 1 is divided by x - 1 (iv) Use remainder theorem to find remainder when x⁴ + x² + 1 is divided by x + 3 (v) Use remainder theorem to find remainder when x⁴ + x³ + 2 is divided by x + 2
Given
(i) Function f(x) = x2 + 5x + 6
Divisor x - 2
Formula
By Remainder Theorem Remainder = f(2)
Working
Substitute x = 2 f(2) = (2)2 + 5(2) + 6
Calculate = 4 + 10 + 6 = 20
Result
Remainder 20
Given
(ii) Function f(x) = x3 + 5x2 + 6
Divisor note x + 1 ⇒ x = -1
Formula
By Remainder Theorem Remainder = f(-1)
Working
Substitute x = -1 f(-1) = (-1)3 + 5(-1)2 + 6
Calculate = -1 + 5 + 6 = 10
Result
Remainder 10
Given
(iii) Function f(x) = x4 + x3 + x2 + x + 1
Divisor note x - 1 ⇒ x = 1
Formula
By Remainder Theorem Remainder = f(1)
Working
Substitute x = 1 f(1) = 1 + 1 + 1 + 1 + 1
Calculate = 5
Result
Remainder 5
Given
(iv) Function f(x) = x4 + x2 + 1
Divisor note x + 3 ⇒ x = -3
Formula
By Remainder Theorem Remainder = f(-3)
Working
Substitute x = -3 f(-3) = (-3)4 + (-3)2 + 1
Calculate = 81 + 9 + 1 = 91
Result
Remainder 91
Given
(v) Function f(x) = x4 + x3 + 2
Divisor note x + 2 ⇒ x = -2
Formula
By Remainder Theorem Remainder = f(-2)
Working
Substitute x = -2 f(-2) = (-2)4 + (-2)3 + 2
Calculate = 16 - 8 + 2 = 10
Result
Remainder 10
3.(i) Use factor theorem to determine if x + 1 is a factor of x² - 1 (ii) Use factor theorem to determine if x - 2 is a factor of x² - 5x + 6 (iii) Use factor theorem to determine if x + 1 is a factor of x³ + x² + x - 3 (iv) Use factor theorem to determine if x - 2 is a factor of x³ + x² - 7x + 2 (v) Use factor theorem to determine if x - 3 is a factor of x⁴ - 3x³ + x² - x + 1
Given
(i) Function f(x) = x2 - 1
Factor to test x + 1 ⇒ x = -1
Formula
By Factor Theorem f(-1) = 0 if x+1 is a factor
Working
Evaluate f(-1) = (-1)2 - 1 = 0
Result
Conclusion x + 1 is a factor
Given
(ii) Function f(x) = x2 - 5x + 6
Factor to test x - 2 ⇒ x = 2
Formula
By Factor Theorem f(2) = 0 if x-2 is a factor
Working
Evaluate f(2) = (2)2 - 5(2) + 6 = 4 - 10 + 6 = 0
Result
Conclusion x - 2 is a factor
Given
(iii) Function f(x) = x3 + x2 + x - 3
Factor to test x + 1 ⇒ x = -1
Formula
By Factor Theorem f(-1) = 0 if x+1 is a factor
Working
Evaluate f(-1) = (-1)3 + (-1)2 + (-1) - 3 = -1 + 1 - 1 - 3 = -4
Result
Conclusion x + 1 is NOT a factor
Given
(iv) Function f(x) = x3 + x2 - 7x + 2
Factor to test x - 2 ⇒ x = 2
Formula
By Factor Theorem f(2) = 0 if x-2 is a factor
Working
Evaluate f(2) = (2)3 + (2)2 - 7(2) + 2 = 8 + 4 - 14 + 2 = 0
Result
Conclusion x - 2 is a factor
Given
(v) Function f(x) = x4 - 3x3 + x2 - x + 1
Factor to test x - 3 ⇒ x = 3
Formula
By Factor Theorem f(3) = 0 if x-3 is a factor
Working
Evaluate f(3) = (3)4 - 3(3)3 + (3)2 - 3 + 1 = 81 - 81 + 9 - 3 + 1 = 7
Result
Conclusion x - 3 is NOT a factor
4.(i) Use synthetic division to show x = 2 is a zero of x³ - 7x + 6 and factorize completely (ii) Use synthetic division to show x = -4 is a zero of x³ - 28x - 48 and factorize completely (iii) Use synthetic division to show x = 2 and x = -3 are zeros of 2x⁴ + 7x³ - 4x² - 27x - 18 and factorize completely
Given
(i) Polynomial f(x) = x3 - 7x + 6
Test value x = 2
Formula
Synthetic division by x - 2 begin{array}{c|cccc} 2 & 1 & 0 & -7 & 6 & & 2 & 4 & -6 hline & 1 & 2 & -3 & 0 end{array}
Quotient polynomial x2 + 2x - 3
Factor the quotient x2 + 2x - 3 = (x+3)(x-1)
Result
Complete factorization x3 - 7x + 6 = (x-2)(x-1)(x+3)
Given
(ii) Polynomial f(x) = x3 - 28x - 48
Test value x = -4
Formula
Synthetic division by x + 4 begin{array}{c|cccc} -4 & 1 & 0 & -28 & -48 & & -4 & 16 & 48 hline & 1 & -4 & -12 & 0 end{array}
Quotient polynomial x2 - 4x - 12
Factor the quotient x2 - 4x - 12 = (x-6)(x+2)
Result
Complete factorization x3 - 28x - 48 = (x+4)(x-6)(x+2)
Given
(iii) Polynomial f(x) = 2x4 + 7x3 - 4x2 - 27x - 18
Test values x = 2, x = -3
Formula
Synthetic division by x - 2 begin{array}{c|ccccc} 2 & 2 & 7 & -4 & -27 & -18 & & 4 & 22 & 36 & 18 hline & 2 & 11 & 18 & 9 & 0 end{array}
Quotient after first division 2x3 + 11x2 + 18x + 9
Synthetic division by x + 3 begin{array}{c|cccc} -3 & 2 & 11 & 18 & 9 & & -6 & -15 & -9 hline & 2 & 5 & 3 & 0 end{array}
Quotient after second division 2x2 + 5x + 3 = (2x+3)(x+1)
Result
Complete factorization 2x4 + 7x3 - 4x2 - 27x - 18 = (x-2)(x+3)(x+1)(2x+3)
5.Use synthetic division to find quotient and remainder when x⁴ - 10x² - 2x + 4 is divided by x + 3
Given
Polynomial f(x) = x4 - 10x2 - 2x + 4
Divisor note x + 3 ⇒ x = -3
Formula
Synthetic division begin{array}{c|ccccc} -3 & 1 & 0 & -10 & -2 & 4 & & -3 & 9 & 3 & -3 hline & 1 & -3 & -1 & 1 & 1 end{array}
Result
Quotient x3 - 3x2 - x + 1
Remainder 1
6.If x + 1 and x - 2 are factors of x³ - px² + qx + 2, use synthetic division to find p and q
Given
Polynomial P(x) = x3 - px2 + qx + 2
Factors x+1, x-2
Formula
Condition from x + 1 being a factor P(-1) = -1 - p - q + 2 = 0 ⇒ -p - q + 1 = 0 ⇒ p + q = 1
Condition from x - 2 being a factor P(2) = 8 - 4p + 2q + 2 = 0 ⇒ 10 - 4p + 2q = 0 ⇒ 2p - q = 5
Solve system p + q = 1 and 2p - q = 5 ⇒ 3p = 6 ⇒ p = 2
Find q 2 + q = 1 ⇒ q = -1
Result
Values p = 2, q = -1
7.When 4x⁴ + 2x³ + kx² + 13 is divided by x + 1, remainder is 16. Find k
Given
Polynomial P(x) = 4x4 + 2x3 + kx2 + 13
Divisor x + 1
Remainder 16
Formula
By Remainder Theorem P(-1) = 16
Working
Evaluate P(-1) P(-1) = 4(-1)4 + 2(-1)3 + k(-1)2 + 13
Simplify = 4 - 2 + k + 13 = k + 15
Solve for k k + 15 = 16 ⇒ k = 1
Result
Value of k k = 1
8.When x³ + x² + x + k is divided by x + 1, remainder is 7. Find k
Given
Polynomial P(x) = x3 + x2 + x + k
Divisor x + 1
Remainder 7
Formula
By Remainder Theorem P(-1) = 7
Working
Evaluate P(-1) P(-1) = (-1)3 + (-1)2 + (-1) + k
Simplify = -1 + 1 - 1 + k = k - 1
Solve for k k - 1 = 7 ⇒ k = 8
Result
Value of k k = 8
9.Use factor theorem to find values of p and q if x + 1 and x - 2 are factors of x³ + px² + qx + 3
Given
Polynomial f(x) = x3 + px2 + qx + 3
Factors x+1, x-2
Formula
Condition from x + 1 f(-1) = -1 + p - q + 3 = 0 ⇒ p - q = -2
Condition from x - 2 f(2) = 8 + 4p + 2q + 3 = 0 ⇒ 4p + 2q = -11
From first equation p = q - 2
Substitute into second 4(q-2) + 2q = -11 ⇒ 6q = -3 ⇒ q = -12
Find p p = -12 - 2 = -52
Result
Values p = -52, q = -12
10.Use factor theorem to find values of a and b if -2 and 2 are roots of 2x³ + 4x² + ax + b
Given
Polynomial f(x) = 2x3 + 4x2 + ax + b
Roots x = -2, x = 2
Formula
Condition from x = -2 f(-2) = 2(-8) + 4(4) + a(-2) + b = 0 ⇒ -16 + 16 - 2a + b = 0 ⇒ b = 2a
Condition from x = 2 f(2) = 2(8) + 4(4) + a(2) + b = 0 ⇒ 16 + 16 + 2a + b = 0 ⇒ 2a + b = -32
Substitute b = 2a 2a + 2a = -32 ⇒ 4a = -32 ⇒ a = -8
Find b b = 2(-8) = -16
Result
Values a = -8, b = -16