Answer :

**Concept Used:**

**Factor theorem: If (x – a) is a factor of f(x), then f(a) = 0**

**Explanation:**

By long division method, we have

**Therefore, the remainder obtained is 5a when x ^{3} – ax^{2} + 6x – a is divided by x – a**

**Remainder theorem:**

Now to find the remainder when x

^{3}– ax

^{2}+ 6x – a is divided by x – a

we have to put x – a = 0, Thus x = a

Let f(x) = x

^{3}– ax

^{2}+ 6x – a

and f(a) will be the remainder

f(a) = a

^{3}– a. a

^{2}+ 6 a – a

f(a) = a

^{3}– a

^{3}+ 5 a

**f(a) = 5 a**

**Hence 5 a is the remainder when x**

^{3}– ax^{2}+ 6x – a is divided by x – a.Rate this question :

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