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# Find the remainder when x^{3} - ax^{2} + 6x - a is divided by x - a.

**Solution:**

Let p(x) be any polynomial of degree greater than or equal to one and let 'a' be any real number. If a polynomial p(x) is divided by x - a then the remainder is p(a).

Let p(x) = x^{3} - ax^{2} + 6x - a

The root of x - a = 0 is a.

p(a) = (a)^{3} - a(a)^{2} + 6(a) - a

= a^{3} - a^{3} + 5a

= 5a

Hence by remainder theorem, 5a is the remainder when x^{3} - ax^{2} + 6x - a is divided by x - a.

**☛ Check: **NCERT Solutions Class 9 Maths Chapter 2

**Video Solution:**

## Find the remainder when x³ - ax² + 6x - a is divided by x - a.

NCERT Solutions Class 9 Maths Chapter 2 Exercise 2.3 Question 2:

**Summary:**

The remainder when x^{3} - ax^{2} + 6x - a is divided by x - a is 5a.

**☛ Related Questions:**

- Verify whether the following are zeroes of the polynomial, indicated against them.(i) p(x) = 3x + 1, x = -(1/3)(ii) p(x) = 5x - π , x = 4/5(iii) p(x) = x2 - 1, x = 1, -1(iv) p(x) = (x + 1)(x - 2), x = -1, 2(v) p(x) = x2 , x = 0(vi) p(x) = lx + m, x = -(m/l)(vii) p(x) = 3x2 - 1, x = -(1/√3), 2/√3(viii) p(x) = 2x + 1, x = 1/2
- Find the zero of the polynomials in each of the following cases:(i) p(x) = x + 5(ii) p(x) = x - 5(iii) p(x) = 2x + 5(iv) p(x) = 3x - 2(v) p(x) = 3x(vi) p(x) = ax, a ≠ 0(vii) p(x) = cx + d, c ≠ 0, c, d are real numbers.
- Find the remainder when x3 + 3x2 + 3x + 1 is divided byi) x + 1ii) x - (1/2)iii) xiv) x + πv) 5 + 2x

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