Check whether 7+3x is a factor of 3x^3+7x ? [ncert]

                              

Now take 7+3x=0 =>3x= -7 => x= -7/3

=> If p(x)=3x^3+7x , then,

=> p(-7/3) = 3(-7/3)^3+7(-7/3)

=> p(-7/3)= 3(-343/27)-49/3

=> p(-7/3)= -343/9 -49/3

=> p(-7/3)=(-343-147)/9

=> p(-7/3)= -490/9 ≠ 0

=> -7/3 is not a zero of 3x^3+7x  or 7+3x is not a factor of 3x^3+7x.

Find the remainder when x^3--ax^2+6x-a is divided by x-a . [ncert]

                           

Let p(x)=x^3-ax^2+6x-a.  - - - - - - (1)

Now put x-a=0 => x=a

Put x=a in (1), we get,

p(a)=a^3-a.a^2+6a-a

p(a)=a^3-a^3+5a

p(a)= 5a

So remainder = 5a Ans.

Find the remainder when x^3+3x^2+3x+1 is divided by x+π. [ncert]

Find the remainder when x^3+3x^2+3x+1 is divided by x-1/2. [ncert]

 

Find the remainder when x^3+3x^2+3x+1 is divided by (i) x+1. [ncert]

Q(3). Verify whether the following are the zeros of the polynomial indicated against them:- (i) p(x)=3x x + 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, x=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 [ncert]

                        Excercise 2.2

Q(3). 

Solution:- We know that if x=α is the zero of a polynomial p(x), then p(α)=0

(i) p(x)= 3x+1, x= -1/3

=> p(-1/3)= 3(-1/3) + 1

=> p(-1/3)= -1+1=0

=> x= -1/3 is a zero of p(x)=3x+1

(ii) p(x)=5x-π , x=4/5

p(x)= 5x - 22/7         [∵ π = 22/7 ]

=> p(4/5) = 5(4/5) - 22/7

=> p(4/5) = 4 - 22/7

=> p(4/5)=  (28-22)/7

=> p(4/5) = 6/7 which is non zero, so x = 4/5 is not a zero of 5x-π

(iii) Do yourself. 

(iv) Do yourself. 

(v) p(x)= x² , x=0

p(0)=0²=0

=> x=0 is a zero of p(x) = x²

(vi) p(x)=lx+m, x= -m/l

=> p(-m/l) = l(-m/l) +m

=> p(-m/l)= -m+m=0

=> x= -m/l is a zero of p(x)= lx+m.

(vii) p(x)=3x²-1, x= -1/√3, 2/√3

p(-1/√3)= 3(-1/√3)²-1

=> p(-1/√3)=3(1/3)-1  

 [Here (-1/√3)²=(-1/√3)(-1/√3)=1/3]

=> p(-1/√3)=1-1=0

=> x= -1/√3 is a zero of p(x)=3x²-1

Now p(2/√3).     Try yourself. 

(viii) p(x)=2x+1, x=1/2.    Try yourself.