Question

Verify whether the following are zeroes of the polynomial, indicated against them.

(i) p(x) = 3x + 1, x =  (ii) p(x) = 5x – π,x = 

(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 = 

(vii) p(x) = 3x2 – 1, x =  (viii)p(x) = 2x + 1, x = 

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Answer 1

Given: Some polynomials with their zeroes.

To find: Verify whether the following are zeroes of the polynomial, indicated against them.

Solution:

• (i) p(x) = 3x + 1, x = -1/3

                 Putting the value of x in equation, we get:

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

                 Yes.

• (ii) p(x) = 5x – π, x =  4/5

                  Putting the value of x in equation, we get:

                 p(4/5) = 5(4/5) - π = 4 - π

                 No

• (iii) p(x) = x^2 – 1, x = 1, –1

                 Putting the value of x in equation, we get:

                 p(1) = 1-1 = 0

                 p(-1) = -1^2 - 1 = 1 - 1 = 0

                 Yes.

• (iv) p(x) = (x + 1) (x – 2), x = – 1, 2

                  Putting the value of x in equation, we get:

                 p(-1) = (-1+1)(-1-2) = 0

                 p(2) = (2+1)(2-2) = 0

                 Yes.

• (v) p(x) = x^2, x = 0

                 Putting the value of x in equation, we get:

                 p(0) = 0^2 = 0

                 Yes.

• (vi) p(x) = lx + m, x =  -m/l

                  Putting the value of x in equation, we get:

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

                 Yes.

• (vii) p(x) = 3x^2 – 1, x = -1/√3, 2/√3

                 Putting the value of x in equation, we get:

                 p(-1/√3) = 3(-1/√3)^2 - 1 = 1 - 1 = 0

                 p(2/√3 ) = 3(2/√3 )^2 - 1 = 4 - 1 = 3

                 -1/√3 is zero of this polynomial but 2/√3 is not.

• (viii)p(x) = 2x + 1, x = 1/2

                 Putting the value of x in equation, we get:

                 p(1/2) = 2(1/2) + 1 = 1 + 1 = 2

                 No.

Answer:

             So the zeroes are verified, (ii) , (vii) and (viii) are not the zeroes of indicated polynomial.

Answer 2

(i)  $x = \frac{-1}{3}$ is the zero of the given polynomial.

(ii)  $x = \frac{4}{5}$  is not the zero of the given polynomial.

(iii)  $x =1,-1$  are the zeroes of the given polynomial.

(iv)  $x =-1,2$  are the zeroes of the given polynomial.

(v) $x =0$  is the zero of the given polynomial.

(vi)  $x =\frac{-m}{l}$  is the zero of the given polynomial.

(vii) $x = \frac{-1}{\sqrt{3} }$ is the zero of the given polynomial and  $x = \frac{2}{\sqrt{3} }$ is not the zero of the given polynomial.

(viii)  $x=\frac{1}{2}$  is not the zero of the given polynomial.

Step-by-step explanation:

We are given with the following polynomials and we have to identify that the values given against them are their zeroes or not.

To identify whether the number is a zero of the given polynomial, put that value in the polynomial equation and if the result comes out to be 0, then that value will be the zero of the given polynomial.

(i) The given polynomial is: p(x) = 3x + 1 ,  $x = \frac{-1}{3}$

Now, put the given value of x in the polynomial;

            $p(\frac{-1}{3}) = 3(\frac{-1}{3})+1$

                       = -1 + 1 = 0

Since the result comes out to be 0, then $x = \frac{-1}{3}$ is the zero of the given polynomial.

(ii) The given polynomial is: p(x) = $5x-\pi$ ,  $x = \frac{4}{5}$

Now, put the given value of x in the polynomial;

            $p(\frac{4}{5}) = 5(\frac{4}{5})-\pi$

                     = $4-\pi$ $\neq$ 0

Since the result doesn't come out to be 0, then $x = \frac{4}{5}$  is not the zero of the given polynomial.

(iii) The given polynomial is: p(x) = $x^{2} -1 , \ x = 1,-1$

Now, first put the given value of x = 1 in the polynomial;

            $p(1) = (1)^{2} -1$

                    = 1 - 1 = 0

Also, put x = -1 in the given polynomial;

            $p(-1) = (-1)^{2} -1$

                    = 1 - 1 = 0

Since the result comes out to be 0, then $x =1,-1$  are the zeroes of the given polynomial.

(iv) The given polynomial is: $p(x) = (x+1)(x-2), \ x =-1,2$

Now, first put the given value of x = -1 in the polynomial;

            $p(-1) = (-1+1)(-1-2)$

                       = $0 \times (-3)$ = 0

Also, put x = 2 in the given polynomial;

            $p(2) = (2+1)(2-2)$

                    = $3 \times 0$ = 0

Since the result comes out to be 0, then $x =-1,2$  are the zeroes of the given polynomial.

(v) The given polynomial is: $p(x)=x^{2} , \ x = 0$

Now, put the given value of x = 0 in the polynomial;

            $p(0)=(0)^{2}$ = 0

Since the result comes out to be 0, then $x =0$  is the zero of the given polynomial.

(vi) The given polynomial is: $p(x)=lx+m, \ x = \frac{-m}{l}$

Now, put the given value of $x=\frac{-m}{l}$ in the polynomial;

            $p(\frac{-m}{l})=l(\frac{-m}{l})+m$

                        = -m + m = 0

Since the result comes out to be 0, then $x =\frac{-m}{l}$  is the zero of the given polynomial.

(vii) The given polynomial is: $p(x) = 3x^{2} -1, \ x = \frac{-1}{\sqrt{3} },\frac{2}{\sqrt{3} }$

Now, first put the given value of $x = \frac{-1}{\sqrt{3} }$ in the polynomial;

            $p(\frac{-1}{\sqrt{3} }) = 3(\frac{-1}{\sqrt{3} })^{2} -1$

                      = $3 \times (\frac{1}{3})-1$

                      = 1 - 1 = 0

Since the result comes out to be 0, then $x = \frac{-1}{\sqrt{3} }$ is the zero of the given polynomial.

Also, put $x = \frac{2}{\sqrt{3} }$ in the given polynomial;

            $p(\frac{2}{\sqrt{3} }) = 3(\frac{2}{\sqrt{3} })^{2} -1$

                      = $3 \times (\frac{4}{3})-1$

                      = 4 - 1 = 3

Since the result doesn't come out to be 0, then  $x = \frac{2}{\sqrt{3} }$ is not the zero of the given polynomial.

(viii) The given polynomial is: $p(x)= 2x+1, \ x = \frac{1}{2}$

Now, put the given value of $x=\frac{1}{2}$ in the polynomial;

            $p(x)= 2(\frac{1}{2})+1$

                    = 1 + 1 = 2

Since the result doesn't come out to be 0, then $x=\frac{1}{2}$  is not the zero of the given polynomial.