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Q1. Verify whether 2 is a zero of the polynomial x³ + 4x² - 3x - 18 or not?

Solution

p(x) = x³ + 4x² - 3x - 18 Now, 2 will be a zero of the polynomial if p(2) = 0. We have: p(2) = (2)³ + 4(2)² - 3(2) - 18        = 8 + 16 - 6 - 18        = 24 - 24        = 0 Thus, 2 is a zero of the polynomial x³ + 4x² - 3x - 18.

Q2. Complete the following. (i) A polynomial of degree three is called a …….polynomial. (ii) A polynomial of degree zero is called a …….polynomial. (iii) A polynomial of degree two is called a …….polynomial. (iv) A polynomial of degree one is called a …….polynomial.

Solution

(i) A polynomial of degree three is called a cubic polynomial.   (ii) A polynomial of degree zero is called a constant polynomial.   (iii) A polynomial of degree two is called a quadratic polynomial.   (iv) A polynomial of degree one is called a linear polynomial.

Q3. What is the geometrical meaning of the zeros of a polynomial?

Solution

Geometrically the zeros of a polynomial f(x) are the x-coordinates of the points where the graph y = f(x) intersects the x-axis.

Q4. Classify the following as linear, bi-quadratic, cubic and constant polynomials.   (i) 4m⁴ + 4m² – 4m - 4    (ii) 8   (iii) 5x + 6x³   -7x² – 2   (iv) 9 - 2y

Solution

(i) 4m⁴ + 4m² – 4m - 4 is bi-quadratic as its highest power is 4. (ii) 8 is a constant as it do not contains a variable. (iii) 5x + 6x³ - 7x² – 2 is a cubic polynomial as its highest power is 3. (iv) 9 - 2y is a linear polynomial as its highest power is 1.

Q5. Find the quotient and remainder when x⁵ +3x⁴ - 5x³ + 14x² + 39x - 11 is divided by 4x + x² - 2.

Solution

Rearranging the terms of divisor in descending order of degree, we get  Clearly, the degree of remainder 9x + 5 is less than the degree of divisor. x² + 4x - 2. ∴ Quotient, q(x) = x³ - x² + x + 8 and remainder, r(x) = 9x + 5.

Q6. If the remainder on division of x³ + 2x² + kx + 3 by x - 3 is 21, then find the quotient and value of k. Hence, find the zeroes of the cubic polynomial x³ + 2x² + kx - 18.

Solution

We have the following terms: Dividend: f(x) = x³ + 2x² + kx + 3, Divisor: g(x) = x - 3 and remainder, r (x) = 21 Using the remainder thermo, we have the following expression: f(3) = 21  The polynomial p(x) is x³ + 2x² - 9x + 3. Now, on long division, we get    Thus, x³ + 2x² - 9x + 3 = (x - 3 ) (x² + 5x + 6) + 21 ∴ The quotient = x² + 5x + 6 Clearly, x³ + 2x² - 9x - 18 = (x - 3 ) (x² + 5x + 6)  = (x - 3 ) (x + 2)(x + 3) Therefore, the zeroes of x³ + 2x² - 9x - 18 are 3, -2 and -3.

Q7. Find the zeroes of the polynomial x² + 3x - 10 and verify the relation between its zeroes and coefficient.  

Solution

We have p(x) = x² + 3x - 10 = (x + 5)(x - 2) For any zero, p(x) = 0  x² + 3x - 10 = 0  (x + 5)(x - 2) = 0  (x + 5) = 0 OR (x - 2) = 0  x = - 5 OR x = 2 The zeros of p(x) = x² + 3x - 10 are as α = -5 and β = 2 Now, sum of zeros = α + β = -5 + 2 = -3 =  Product of zeros = αβ = (-5) × 2 = - 10 =     

Q8. Find the quotient and reminder using division algorithm: f(x) = x³ - 6x² + 11x - 6, g(x) = x + 1  

Solution

We have: f(x) = x³ - 6x² + 11x - 6 and g(x) = x + 1 Clearly, degree of f(x) = 3 and degree of g(x) = 1. Therefore, the degree of quotient is q(x) = 3 - 1 = 2 and the degree of remainder is r(x) = 0 Let quotient q(x) = ax² + bx + c and remainder r(x) = k. Using division algorithm, we have f(x) = g(x) × q(x) + r(x)   Comparing the coefficient of same powers of x on both sides, we get a = 1  [Comparing the coefficient of x³] a + b = -6  [Comparing the coefficient of x²] b + c = 11  [Comparing the coefficient of x] c + k = -6  [Comparing the constant terms ] Solving the above equations, we get the following values: a = 1, b = -7, c = 18, and k = -24 ∴ Quotient is q (x) = x² - 7x + 18 and remainder is r(x) = -24.        

Q9. Find the quotient and reminder using division algorithm: f(x) = 10x⁴ + 17x³ - 62x² + 30x - 3, g(x)=2x² + 7x + 1  

Solution

We have the following equations: f(x) = 10x⁴ + 17x³ - 62x² + 30x - 3 and g(x) = 2x² + 7x + 1. Clearly, degree of f(x) = 4 and degree of g(x) = 2. Therefore, degree of quotient, q(x) = 4 - 2 = 2 and degree of remainder, r(x) is less than degree of g(x), i.e. 2. Let quotient, q(x) = ax² + bx + c and remainder, r(x) = dx + e Using division algorithm, we have the following equation: f(x) = g(x) × q(x) + r(x) ∴10x⁴ + 17x³ - 62x² + 30x - 3 = (2x² + 7x + 1) (ax² + bx + c) + (dx + e) = 2ax⁴ + 2bx³ + 2cx² + 7ax³ + 7bx² + 7cx + ax² + bx + c + dx + e = 2ax⁴ + (7a + 2b)x³ + (a + 7b + 2c)x² + (b + 7c + d)x + (c + e) Comparing the coefficient of same powers of x on both sides, we get 2a = 10  [Comparing the power of x⁴] 7a + 2b = 17 [Comparing the power of x³] a + 7b + 2c = -62 [comparing the power of x²] b + 7c + d = 30  [comparing the power of x] c + e = -3  [comparing the constant terms] Solving the above equations, we get the following values: a = 5, b = -9, c = -2, d = 53 and e = -1 Hence, quotient, q (x) = 5x² - 9x - 4 and remainder, r(x) = 53x - 1.   

Q10. Find the quotient and reminder using division algorithm: f(x) = 15x³ - 20x² + 13x - 12, g(x) = 2 - 2x + x²  

Solution

We have the following equations: f(x) = 15x³ - 20x² + 13x - 12 and g(x) = 2 - 2x + x² is x² - 2x + 2 Clearly, degree of f(x) = 3 and degree of g(x) = 2. Therefore, the degree of quotient q(x) = 3 - 2 = 1 and degree of remainder r(x) is less than the degree of g(x), i.e. 2. Let quotient, q(x) = ax + b and remainder, r(x) = cx + d. Using division algorithm, we have the following equation: p(x) = g(x) × q(x) + r(x) ∴15x³ - 20x² + 13x - 12 = (x² - 2x + 2) (ax + b) + (cx + d) = ax³ + bx² - 2ax² - 2bx + 2ax + 2b + cx + d = ax³ + (b - 2a)x² + (2a + c - 2b) x + (2b + d) Comparing the coefficient of same powers of x on both sides, we get a = 15  [Comparing the coefficient of x³] b - 2a = -20  [Comparing the coefficient of x²] 2a + c - 2b = 13  [Comparing the coefficient of x] 2b + d = -12  [Comparing the constant terms ] Solving the above equation, we get the following values: a = 15, b = 10, c = 3 and d = -32 Hence, quotient, q(x) = 15x + 10 and remainder, r(x) = 3x - 32.