Page 1 Exercise 2B 1. Verify that 3, -2, 1 are the zeros of the cubic polynomial p(x) = (x 3 2x 2 5x + 6) and verify the relation between it zeros and coefficients. Sol: The given polynomial is p(x) = (x 3 2x 2 5x + 6) p(3) = (3 3 2 × 3 2 5 × 3 + 6) = (27 18 15 + 6) = 0 p(-2) = [ ( 2 3 ) 2 × ( 2) 2 5 × ( 2) + 6] = ( 8 8 + 10 + 6) = 0 p(1) = (1 3 2 × 1 2 5 × 1 + 6) = ( 1 2 5 + 6) = 0 3, 2 and 1are the zeroes of p(x), Let = 3, = 2 and = 1. Then we have: ( ) = (3 2 + 1) = 2 = ( ) = ( 6 2 + 3) = = = { 3 × (-2) × 1} = = 2. Verify that 5, -2 and are the zeroes of the cubic polynomial p(x) = (3x 3 10x 2 27x + 10) and verify the relation between its zeroes and coefficients. Sol: p(x) = (3x 3 10x 2 27x + 10) p(5) = (3 × 5 3 10 × 5 2 27 × 5 + 10) = (375 250 135 + 10) = 0 p( 2) = [3 × ( 2 3 ) 10 × ( 2 2 ) 27 × ( 2) + 10] = ( 24 40 + 54 + 10) = 0 p = = = = = = 0 5, 2 and are the zeroes of p(x). Let = 5, = 2 and = . Then we have: ( ) = = = ( ) = = = = = = Page 2 Exercise 2B 1. Verify that 3, -2, 1 are the zeros of the cubic polynomial p(x) = (x 3 2x 2 5x + 6) and verify the relation between it zeros and coefficients. Sol: The given polynomial is p(x) = (x 3 2x 2 5x + 6) p(3) = (3 3 2 × 3 2 5 × 3 + 6) = (27 18 15 + 6) = 0 p(-2) = [ ( 2 3 ) 2 × ( 2) 2 5 × ( 2) + 6] = ( 8 8 + 10 + 6) = 0 p(1) = (1 3 2 × 1 2 5 × 1 + 6) = ( 1 2 5 + 6) = 0 3, 2 and 1are the zeroes of p(x), Let = 3, = 2 and = 1. Then we have: ( ) = (3 2 + 1) = 2 = ( ) = ( 6 2 + 3) = = = { 3 × (-2) × 1} = = 2. Verify that 5, -2 and are the zeroes of the cubic polynomial p(x) = (3x 3 10x 2 27x + 10) and verify the relation between its zeroes and coefficients. Sol: p(x) = (3x 3 10x 2 27x + 10) p(5) = (3 × 5 3 10 × 5 2 27 × 5 + 10) = (375 250 135 + 10) = 0 p( 2) = [3 × ( 2 3 ) 10 × ( 2 2 ) 27 × ( 2) + 10] = ( 24 40 + 54 + 10) = 0 p = = = = = = 0 5, 2 and are the zeroes of p(x). Let = 5, = 2 and = . Then we have: ( ) = = = ( ) = = = = = = ______________________________________________________________________________ 3. Find a cubic polynomial whose zeroes are 2, -3and 4. Sol: If the zeroes of the cubic polynomial are a, b and c then the cubic polynomial can be found as x 3 (a + b + c)x 2 + (ab + bc + ca)x Let a = 2, b = 3 and c = 4 Substituting the values in 1, we get x 3 (2 3 + 4)x 2 + ( 6 12 + 8)x ( 24) x 3 3x 2 10x + 24 4. Find a cubic polynomial whose zeroes are , 1 and 3. Sol: If the zeroes of the cubic polynomial are a, b and c then the cubic polynomial can be found as x 3 (a + b + c)x 2 + (ab + bc + ca)x Let a = , b = 1 and c = 3 Substituting the values in (1), we get x 3 x 2 + x x 3 x 2 4x + 2x 3 +3x 2 8x + 3 5. Find a cubic polynomial with the sum of its zeroes, sum of the products of its zeroes taken two at a time and the product of its zeroes as 5, -2 and -24 respectively. Sol: We know the sum, sum of the product of the zeroes taken two at a time and the product of the zeroes of a cubic polynomial then the cubic polynomial can be found as x 3 (sum of the zeroes)x 2 + (sum of the product of the zeroes taking two at a time)x product of zeroes Therefore, the required polynomial is x 3 5x 2 2x + 24 6. If f(x) = 3 3 5 3 x x x is divided by g(x) = 2 2 x Sol: x 3 x 2 x 3 3x 2 + 5x 3 x 3 2x - + 3x 2 + 7x 3 3x 2 + 6 + 7x 9 Quotient q(x) = x 3 Remainder r(x) = 7x 9 Page 3 Exercise 2B 1. Verify that 3, -2, 1 are the zeros of the cubic polynomial p(x) = (x 3 2x 2 5x + 6) and verify the relation between it zeros and coefficients. Sol: The given polynomial is p(x) = (x 3 2x 2 5x + 6) p(3) = (3 3 2 × 3 2 5 × 3 + 6) = (27 18 15 + 6) = 0 p(-2) = [ ( 2 3 ) 2 × ( 2) 2 5 × ( 2) + 6] = ( 8 8 + 10 + 6) = 0 p(1) = (1 3 2 × 1 2 5 × 1 + 6) = ( 1 2 5 + 6) = 0 3, 2 and 1are the zeroes of p(x), Let = 3, = 2 and = 1. Then we have: ( ) = (3 2 + 1) = 2 = ( ) = ( 6 2 + 3) = = = { 3 × (-2) × 1} = = 2. Verify that 5, -2 and are the zeroes of the cubic polynomial p(x) = (3x 3 10x 2 27x + 10) and verify the relation between its zeroes and coefficients. Sol: p(x) = (3x 3 10x 2 27x + 10) p(5) = (3 × 5 3 10 × 5 2 27 × 5 + 10) = (375 250 135 + 10) = 0 p( 2) = [3 × ( 2 3 ) 10 × ( 2 2 ) 27 × ( 2) + 10] = ( 24 40 + 54 + 10) = 0 p = = = = = = 0 5, 2 and are the zeroes of p(x). Let = 5, = 2 and = . Then we have: ( ) = = = ( ) = = = = = = ______________________________________________________________________________ 3. Find a cubic polynomial whose zeroes are 2, -3and 4. Sol: If the zeroes of the cubic polynomial are a, b and c then the cubic polynomial can be found as x 3 (a + b + c)x 2 + (ab + bc + ca)x Let a = 2, b = 3 and c = 4 Substituting the values in 1, we get x 3 (2 3 + 4)x 2 + ( 6 12 + 8)x ( 24) x 3 3x 2 10x + 24 4. Find a cubic polynomial whose zeroes are , 1 and 3. Sol: If the zeroes of the cubic polynomial are a, b and c then the cubic polynomial can be found as x 3 (a + b + c)x 2 + (ab + bc + ca)x Let a = , b = 1 and c = 3 Substituting the values in (1), we get x 3 x 2 + x x 3 x 2 4x + 2x 3 +3x 2 8x + 3 5. Find a cubic polynomial with the sum of its zeroes, sum of the products of its zeroes taken two at a time and the product of its zeroes as 5, -2 and -24 respectively. Sol: We know the sum, sum of the product of the zeroes taken two at a time and the product of the zeroes of a cubic polynomial then the cubic polynomial can be found as x 3 (sum of the zeroes)x 2 + (sum of the product of the zeroes taking two at a time)x product of zeroes Therefore, the required polynomial is x 3 5x 2 2x + 24 6. If f(x) = 3 3 5 3 x x x is divided by g(x) = 2 2 x Sol: x 3 x 2 x 3 3x 2 + 5x 3 x 3 2x - + 3x 2 + 7x 3 3x 2 + 6 + 7x 9 Quotient q(x) = x 3 Remainder r(x) = 7x 9 7. If f(x) = x 4 3x 2 + 4x + 5 is divided by g(x)= x 2 x + 1 Sol: x 2 + x 3 x 2 x + 1 x 4 + 0x 3 3x 2 + 4x + 5 x 4 x 3 + x 2 - + x 3 4x 2 + 4x + 5 x 3 x 2 + x + 3x 2 + 3x + 5 3x 2 + 3x 3 + + 8 Quotient q(x) = x 2 + x 3 Remainder r(x) = 8 8. If f(x) = x 4 5x + 6 is divided by g(x) = 2 x 2 . Sol: We can write f(x) as x 4 + 0x 3 + 0x 2 5x + 6 and g(x) as x 2 + 2 x 2 2 x 2 + 2 x 4 + 0x 3 + 0x 2 5x + 6 x 4 2 x 2 - + 2x 2 5x + 6 2x 2 4 + 5x + 10 Quotient q(x) = x 2 2 Remainder r(x) = 5x + 10 Page 4 Exercise 2B 1. Verify that 3, -2, 1 are the zeros of the cubic polynomial p(x) = (x 3 2x 2 5x + 6) and verify the relation between it zeros and coefficients. Sol: The given polynomial is p(x) = (x 3 2x 2 5x + 6) p(3) = (3 3 2 × 3 2 5 × 3 + 6) = (27 18 15 + 6) = 0 p(-2) = [ ( 2 3 ) 2 × ( 2) 2 5 × ( 2) + 6] = ( 8 8 + 10 + 6) = 0 p(1) = (1 3 2 × 1 2 5 × 1 + 6) = ( 1 2 5 + 6) = 0 3, 2 and 1are the zeroes of p(x), Let = 3, = 2 and = 1. Then we have: ( ) = (3 2 + 1) = 2 = ( ) = ( 6 2 + 3) = = = { 3 × (-2) × 1} = = 2. Verify that 5, -2 and are the zeroes of the cubic polynomial p(x) = (3x 3 10x 2 27x + 10) and verify the relation between its zeroes and coefficients. Sol: p(x) = (3x 3 10x 2 27x + 10) p(5) = (3 × 5 3 10 × 5 2 27 × 5 + 10) = (375 250 135 + 10) = 0 p( 2) = [3 × ( 2 3 ) 10 × ( 2 2 ) 27 × ( 2) + 10] = ( 24 40 + 54 + 10) = 0 p = = = = = = 0 5, 2 and are the zeroes of p(x). Let = 5, = 2 and = . Then we have: ( ) = = = ( ) = = = = = = ______________________________________________________________________________ 3. Find a cubic polynomial whose zeroes are 2, -3and 4. Sol: If the zeroes of the cubic polynomial are a, b and c then the cubic polynomial can be found as x 3 (a + b + c)x 2 + (ab + bc + ca)x Let a = 2, b = 3 and c = 4 Substituting the values in 1, we get x 3 (2 3 + 4)x 2 + ( 6 12 + 8)x ( 24) x 3 3x 2 10x + 24 4. Find a cubic polynomial whose zeroes are , 1 and 3. Sol: If the zeroes of the cubic polynomial are a, b and c then the cubic polynomial can be found as x 3 (a + b + c)x 2 + (ab + bc + ca)x Let a = , b = 1 and c = 3 Substituting the values in (1), we get x 3 x 2 + x x 3 x 2 4x + 2x 3 +3x 2 8x + 3 5. Find a cubic polynomial with the sum of its zeroes, sum of the products of its zeroes taken two at a time and the product of its zeroes as 5, -2 and -24 respectively. Sol: We know the sum, sum of the product of the zeroes taken two at a time and the product of the zeroes of a cubic polynomial then the cubic polynomial can be found as x 3 (sum of the zeroes)x 2 + (sum of the product of the zeroes taking two at a time)x product of zeroes Therefore, the required polynomial is x 3 5x 2 2x + 24 6. If f(x) = 3 3 5 3 x x x is divided by g(x) = 2 2 x Sol: x 3 x 2 x 3 3x 2 + 5x 3 x 3 2x - + 3x 2 + 7x 3 3x 2 + 6 + 7x 9 Quotient q(x) = x 3 Remainder r(x) = 7x 9 7. If f(x) = x 4 3x 2 + 4x + 5 is divided by g(x)= x 2 x + 1 Sol: x 2 + x 3 x 2 x + 1 x 4 + 0x 3 3x 2 + 4x + 5 x 4 x 3 + x 2 - + x 3 4x 2 + 4x + 5 x 3 x 2 + x + 3x 2 + 3x + 5 3x 2 + 3x 3 + + 8 Quotient q(x) = x 2 + x 3 Remainder r(x) = 8 8. If f(x) = x 4 5x + 6 is divided by g(x) = 2 x 2 . Sol: We can write f(x) as x 4 + 0x 3 + 0x 2 5x + 6 and g(x) as x 2 + 2 x 2 2 x 2 + 2 x 4 + 0x 3 + 0x 2 5x + 6 x 4 2 x 2 - + 2x 2 5x + 6 2x 2 4 + 5x + 10 Quotient q(x) = x 2 2 Remainder r(x) = 5x + 10 9. By actual division, show that x 2 3 is a factor of 2x 4 + 3x 3 2x 2 9x 12. Sol: Let f(x) = 2x 4 + 3x 3 2x 2 9x 12 and g(x) as x 2 3 2x 2 + 3x + 4 x 2 3 2x 4 + 3x 3 2x 2 9x 12 2x 4 6x 2 - + 3x 3 + 4x 2 9x 12 3x 3 9x + 4x 2 12 4x 2 12 + x Quotient q(x) = 2x 2 + 3x + 4 Remainder r(x) = 0 Since, the remainder is 0. Hence, x 2 3 is a factor of 2x 4 + 3x 3 2x 2 9x 12 10. On dividing 3x 3 + x 2 + 2x + 5 is divided by a polynomial g(x), the quotient and remainder are (3x 5) and (9x + 10) respectively. Find g(x). Sol: By using division rule, we have Dividend = Quotient × Divisor + Remainder 3x 3 + x 2 + 2x + 5 = (3x 5)g(x) + 9x + 10 3x 3 + x 2 + 2x + 5 9x 10 = (3x 5)g(x) 3x 3 + x 2 7x 5 = (3x 5)g(x) g(x) = x 2 + 2x + 1 3x 5 3x 3 + x 2 7x 5 3x 3 5x 2 + 6x 2 7x 5 6x 2 10x + 3x 5 3x 5 + X g(x) = x 2 + 2x + 1 Page 5 Exercise 2B 1. Verify that 3, -2, 1 are the zeros of the cubic polynomial p(x) = (x 3 2x 2 5x + 6) and verify the relation between it zeros and coefficients. Sol: The given polynomial is p(x) = (x 3 2x 2 5x + 6) p(3) = (3 3 2 × 3 2 5 × 3 + 6) = (27 18 15 + 6) = 0 p(-2) = [ ( 2 3 ) 2 × ( 2) 2 5 × ( 2) + 6] = ( 8 8 + 10 + 6) = 0 p(1) = (1 3 2 × 1 2 5 × 1 + 6) = ( 1 2 5 + 6) = 0 3, 2 and 1are the zeroes of p(x), Let = 3, = 2 and = 1. Then we have: ( ) = (3 2 + 1) = 2 = ( ) = ( 6 2 + 3) = = = { 3 × (-2) × 1} = = 2. Verify that 5, -2 and are the zeroes of the cubic polynomial p(x) = (3x 3 10x 2 27x + 10) and verify the relation between its zeroes and coefficients. Sol: p(x) = (3x 3 10x 2 27x + 10) p(5) = (3 × 5 3 10 × 5 2 27 × 5 + 10) = (375 250 135 + 10) = 0 p( 2) = [3 × ( 2 3 ) 10 × ( 2 2 ) 27 × ( 2) + 10] = ( 24 40 + 54 + 10) = 0 p = = = = = = 0 5, 2 and are the zeroes of p(x). Let = 5, = 2 and = . Then we have: ( ) = = = ( ) = = = = = = ______________________________________________________________________________ 3. Find a cubic polynomial whose zeroes are 2, -3and 4. Sol: If the zeroes of the cubic polynomial are a, b and c then the cubic polynomial can be found as x 3 (a + b + c)x 2 + (ab + bc + ca)x Let a = 2, b = 3 and c = 4 Substituting the values in 1, we get x 3 (2 3 + 4)x 2 + ( 6 12 + 8)x ( 24) x 3 3x 2 10x + 24 4. Find a cubic polynomial whose zeroes are , 1 and 3. Sol: If the zeroes of the cubic polynomial are a, b and c then the cubic polynomial can be found as x 3 (a + b + c)x 2 + (ab + bc + ca)x Let a = , b = 1 and c = 3 Substituting the values in (1), we get x 3 x 2 + x x 3 x 2 4x + 2x 3 +3x 2 8x + 3 5. Find a cubic polynomial with the sum of its zeroes, sum of the products of its zeroes taken two at a time and the product of its zeroes as 5, -2 and -24 respectively. Sol: We know the sum, sum of the product of the zeroes taken two at a time and the product of the zeroes of a cubic polynomial then the cubic polynomial can be found as x 3 (sum of the zeroes)x 2 + (sum of the product of the zeroes taking two at a time)x product of zeroes Therefore, the required polynomial is x 3 5x 2 2x + 24 6. If f(x) = 3 3 5 3 x x x is divided by g(x) = 2 2 x Sol: x 3 x 2 x 3 3x 2 + 5x 3 x 3 2x - + 3x 2 + 7x 3 3x 2 + 6 + 7x 9 Quotient q(x) = x 3 Remainder r(x) = 7x 9 7. If f(x) = x 4 3x 2 + 4x + 5 is divided by g(x)= x 2 x + 1 Sol: x 2 + x 3 x 2 x + 1 x 4 + 0x 3 3x 2 + 4x + 5 x 4 x 3 + x 2 - + x 3 4x 2 + 4x + 5 x 3 x 2 + x + 3x 2 + 3x + 5 3x 2 + 3x 3 + + 8 Quotient q(x) = x 2 + x 3 Remainder r(x) = 8 8. If f(x) = x 4 5x + 6 is divided by g(x) = 2 x 2 . Sol: We can write f(x) as x 4 + 0x 3 + 0x 2 5x + 6 and g(x) as x 2 + 2 x 2 2 x 2 + 2 x 4 + 0x 3 + 0x 2 5x + 6 x 4 2 x 2 - + 2x 2 5x + 6 2x 2 4 + 5x + 10 Quotient q(x) = x 2 2 Remainder r(x) = 5x + 10 9. By actual division, show that x 2 3 is a factor of 2x 4 + 3x 3 2x 2 9x 12. Sol: Let f(x) = 2x 4 + 3x 3 2x 2 9x 12 and g(x) as x 2 3 2x 2 + 3x + 4 x 2 3 2x 4 + 3x 3 2x 2 9x 12 2x 4 6x 2 - + 3x 3 + 4x 2 9x 12 3x 3 9x + 4x 2 12 4x 2 12 + x Quotient q(x) = 2x 2 + 3x + 4 Remainder r(x) = 0 Since, the remainder is 0. Hence, x 2 3 is a factor of 2x 4 + 3x 3 2x 2 9x 12 10. On dividing 3x 3 + x 2 + 2x + 5 is divided by a polynomial g(x), the quotient and remainder are (3x 5) and (9x + 10) respectively. Find g(x). Sol: By using division rule, we have Dividend = Quotient × Divisor + Remainder 3x 3 + x 2 + 2x + 5 = (3x 5)g(x) + 9x + 10 3x 3 + x 2 + 2x + 5 9x 10 = (3x 5)g(x) 3x 3 + x 2 7x 5 = (3x 5)g(x) g(x) = x 2 + 2x + 1 3x 5 3x 3 + x 2 7x 5 3x 3 5x 2 + 6x 2 7x 5 6x 2 10x + 3x 5 3x 5 + X g(x) = x 2 + 2x + 1 11. Verify division algorithm for the polynomial f(x)= (8 + 20x + x 2 6x 3 ) by g(x) =( 2 + 5x 3x 2 ). Sol: We can write f(x) as 6x 3 + x 2 + 20x + 8 and g(x) as 3x 2 + 5x + 2 x 2 + 2x + 1 3x 2 + 5x + 2 6x 3 + x 2 + 20x + 8 6x 3 +10x 2 + 4x + 9x 2 +16x + 8 9x 2 +15x + 6 + x + 2 Quotient = 2x + 3 Remainder = x + 2 By using division rule, we have Dividend = Quotient × Divisor + Remainder 6x 3 + x 2 + 20x + 8 = ( 3x 2 + 5x + 2) (2x + 3) + x + 2 6x 3 + x 2 + 20x + 8 = 6x 3 + 10x 2 + 4x 9x 2 + 15x + 6 + x + 2 6x 3 + x 2 + 20x + 8 = 6x 3 + x 2 + 20x + 8 12. It is given that 1 is one of the zeroes of the polynomial x 3 + 2x 2 11x 12. Find all the zeroes of the given polynomial. Sol: Let f(x) = x 3 + 2x 2 11x 12 Since 1 is a zero of f(x), (x+1) is a factor of f(x). On dividing f(x) by (x+1), we get x + 1 x 3 + 2x 2 11x 12 x 2 + x + 12 x 3 + x 2 x 2 11x 12 x 2 + x 12x 12 12x 12 + + XRead More

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