Q1 of 274 Page 55

is divided by (x-1)

Let, f(x) = x3 – 6x2 + 9x + 3


Now,


As per the question,


x – 1 = 0


x = 1


Using Remainder theorem,


We know that when f(x)is divided by (x – 1), the remainder so obtained will be f(1).


Hence,


f(1) = (1)3 – 6(1)2 + 9(1) + 3


= 1 – 6 + 9 + 3


= 13 – 6


= 7


Therefore,


The required remainder is 7


More from this chapter

All 274 →
4

Find the zero of the polynomial:

(i) P(x)=x-5


(ii) q(x)=x+4


(iii) p(t)=2t-3


(iv) f(x)=3x+1


(v) g(x)=5-4x


(vi) h(x)=6x-1


(vii) p(x)=ax + b,a≠0


(viii) q(x) = 4x


(ix) p(x)=ax,a≠0

 5

Verify that:


(i) 4 is a zero of the polynomial p(x)=x-4.


(ii) -3 is a zero of the polynomial p(x) = x + 3.


(iii) -1/2 is a zero of the polynomial p(y)=2y+1.


(iv) 2/5 is a zero of the polynomial p(x) =2-5x.


(v) 1 and 3 are the zeros of the polynomial p(x)=(x-1)(x-2)


(vi) 0 and 3 are the zeros of the polynomial p(x) =x2 - 3x


(vii) 2 and-3 are the zeros of the polynomial p(x) =x2 + x - 6

 2

is divided by (x-3)

 3

is divided by (x-2)