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# If the polynomial x^{4} - 6x^{3} + 16x^{2} - 25x + 10 is divided by another polynomial x^{2} - 2x + k, the remainder comes out to be x + a, find k and a.

**Solution:**

The given polynomial is p(x) = x^{4} - 6x^{3} + 16x^{2} - 25x + 10

We can solve this by using division algorithm i.e, Dividend = Divisor × Quotient + Remainder

Let us divide and equate the obtained remainder with (x + a)

Now, it is given that p(x) when divided by x^{2 }– 2x + k leaves (*x *+ a) as remainder.

Let us equate the remainder with x + a (as given in the question)

(-9 + 2k)x + 10 - 8k + k^{2} = x + a

Let us compare the coefficient of both LHS and RHS.

-9 + 2k = 1

⇒ 2k = 10

⇒ k = 5 ----- (1)

Also, 10 - 8k + k^{2} = a ----- (2)

As we obtained the value of k, let us substitute in equation (2) to find the value of a.

a = 10 - 40 + 25

a = - 5

Therefore, the value of k is 5 and a is - 5.

**☛ Check: **NCERT Solutions Class 10 Maths Chapter 2

**Video Solution:**

## If the polynomial x⁴ - 6x³ + 16x² - 25x + 10 is divided by another polynomial x² - 2x + k, the remainder comes out to be x + a, find k and a.

NCERT Solutions Class 10 Maths Chapter 2 Exercise 2.4 Question 5

**Summary:**

If the polynomial x^{4} - 6x^{3} + 16x^{2} - 25x + 10 is divided by another polynomial x^{2} - 2x + k, the remainder comes out to be x + a. The values of k and a are 5 and - 5 respectively.

**☛ Related Questions:**

- 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) 2x3 + x2 - 5x + 2; 1/2, 1, - 2(ii) x3 - 4x2 + 5x - 2; 2, 1, 1
- 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.
- If the zeroes of the polynomial x^3 - 3x^2 + x + 1 are a - b, a, a + b, find a and b.
- If two zeroes of the polynomial x^4 - 6x^3 - 26x^2 + 138x - 35 are 2 ± √3 find other zeroes.

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