### Polynomials Class 10th Mathematics CBSE Solution

##### Question 1.Verify that the numbers given alongside of the cubic polynomials below are their zeroes. Also verify the relationship between the zeroes and the coefficients in each case:(i) (ii) Answer:(i) P(x) = Now for zeroes, putting the given values in x.P(1/2) = 2(1/2)3 + (1/2)2 - 5(1/2) + 2= (1/4) + (1/4) - (5/2) + 2= (1 + 1 - 10 + 8)/2= 0/2 = 0P(1) = P(-2) = Thus, 1/2, 1 and -2 are zeroes of given polynomial.Comparing given polynomial with ax3 + bx2 + cx + d and Taking zeroes as α, β, and γ, we have Now, We know the relation between zeroes and the coefficient of a standard cubic polynomial as Substituting value, we have  Since, LHS = RHS (Relation Verified)    Since LHS = RHS, Relation verified.   Since LHS = RHS, Relation verified.Thus, all three relationships between zeroes and the coefficient is verified.(ii) p(x) = x3 – 4x2 + 5x – 2Now for zeroes , put the given value in x.P(2) = = P(1) = P(1) = Thus, 2, 1 , 1 are the zeroes of the given polynomial.Now,Comparing the given polynomial with ax3 + bx2 + cx + d, we get Now,  4 = 4   5 = 5αβγ = 2 × 1 × 1 = 22 = 2Thus, all three relationships between zeroes and the coefficient is verified.Question 2.Find a cubic polynomial with the sum, sum of the product of its zeroes taken two at a time, and the product of its zeroes as 2, -7, -14 respectively.Answer:For a cubic polynomial equation, ax3 + bx2 + cx + d, and zeroes α, β and γwe know that Let the polynomial be ax3 + bx2 + cx + d, and zeroes α, β and γ.A cubic polynomial with respect to its zeroes is given by,x3 - (sum of zeroes) x2 + (Sum of the product of roots taken two at a time) x - Product of Roots = 0x3 - (2) x2 + (- 7) x - (- 14) = 0x3 - (2) x2 + (- 7) x + 14 = 0Hence, the polynomial is x3 - 2x2 - 7x + 14.Question 3.If the zeroes of the polynomial are (a - b), a and (a + b). Find a and b.Answer:Given,p(x) = zeroes are = a – b , a + b , aC = Sum of zeroes =  a = The zeroes are = (1 - b), 1 and (1 + b)Product of zeroes = (1 - b)(1 + b)(1 - b)(1 + b) = -q/m   So,We get, Question 4.If two zeroes of the polynomial x4 - 6x3 - 26x2 + 138x - 35 are find other zeroes.Answer:Given:2+√3 and 2-√3 are zeroes of given equation, Therefore, (x - 2 + √3)(x - 2 - √3) should be a factor of given equation.Also, (x - 2 + √3)(x - 2 - √3) = x2 - 2x - √3x -2x + 4 + 2√3 + √3x - 2√3 - 3= x2 - 4x + 1To find other zeroes, we divide given equation by x2 - 4x + 1 We get ,x4 - 6x3 - 26x2 + 138x - 35 = (x2 - 4x + 1)(x2 - 2x - 35)Now factorizing x2 - 2 x - 35 we get, The value of polynomial is also zero when ,x - 7 = 0orx = 7And, x + 5 = 0orx = -5Hence, 7 and -5 are also zeroes of this polynomial.Question 5.If the polynomial x4 - 6x3 + 16x2 - 25x + 10 is divided by another polynomial x2 - 2k + k the remainder comes out to be x + a, find k and a. Answer:To solve this question divide x4 - 6 x3 + 16 x2 - 25 x + 10 by x2 - 2 x + k by long division methodLet us divide, by  So, remainder = (2k - 9)x + (10 - 8k + k2)But given remainder = x + a⇒ (2k - 9)x + (10 - 8k + k2) = x + aComparing coefficient of x, we have2k - 9 = 1⇒ 2k = 10⇒ k = 5andComparing constant term,10 - 8k + k2 = a⇒ a = 10 - 8(5) + 52⇒ a = 10 - 40 + 25⇒ a = -5So, the value of k is 5 and a is -5.

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