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Question

Verify whether the following are zeroes of the polynomial, indicated against them.
(i) $$p(x)=3x+1, x=-\displaystyle\frac{1}{3}$$
(ii) $$p(x)=5x-\pi, x=\displaystyle\frac{4}{5}$$
(iii) $$p(x)=x^2-1, x= 1, -1$$
(iv) $$p(x)=(x+1)(x-2), x=-1, 2$$
(v) $$p(x)=x^2, x= 0$$
(vi) $$p(x)=lx+m, x=-\displaystyle\frac{m}{l}$$
(vii) $$p(x)=3x^2-1, x=-\displaystyle\frac{1}{\sqrt 3},\frac{2}{\sqrt 3}$$
(viii) $$p(x)=2x+1, x=\displaystyle\frac{1}{2}$$


Solution

In order to verify the values are zeros of polynomial $$p(x)$$, we must replace the variable $$x$$ with the given values.
If $$p(x) = 0$$, then that given value is zero of polynomial $$p(x)$$.

(i) $$p(x) = 3x+1$$:
Put $$x=-\dfrac 13$$, we get,
$$p(x) = p(\dfrac{-1}{3}) =3\left(-\dfrac 13\right)+1 = -1+1 = 0$$.
So, $$x=-\dfrac 13$$ is the zero of the given polynomial $$p(x)$$.

(ii) $$p(x) = 5x-\pi$$:
Put $$x=\dfrac 45$$, we get,
$$p(x) = p(\dfrac45)=5\left(\dfrac 45\right)-\pi = 4-\pi \neq 0$$.
So, $$x=\dfrac 45$$ is not the zero of the given polynomial $$p(x)$$.

(iii) $$p(x) = x^2-1$$:
Put $$x=1$$, we get,
$$p(x) =p(1)= (1)^2-1 = 1-1 = 0$$.
So, $$x=1$$ is the zero of the given polynomial $$p(x)$$.
Now put $$x=-1$$, we get,
$$p(x) p(-1)= (-1)^2-1 = 1-1 = 0$$.
So, $$x=-1$$ is the zero of the given polynomial $$p(x)$$.

(iv) $$p(x) = (x+1)(x-2)$$:
Put $$x=-1$$, we get,
$$p(x) = (-1+1)(-1-2)= 0(-3) = 0$$.
So, $$x=-1$$ is the zero of the given polynomial $$p(x)$$.    
Now put $$x=2$$, we get,
$$p(x) = (2+1)(2-2)= 3(0) = 0$$.
So, $$x=2$$ is the zero of the given polynomial $$p(x)$$.

(v) $$p(x) = x^2$$:
Put $$x=0$$, we get,
$$p(x)=p(0)=(0)^2=0$$.
So, $$x=0$$ is the zero of the given polynomial $$p(x)$$.  

(vi) $$p(x) = lx+m$$:
Put $$x=-\dfrac ml$$, we get,
$$p(x) = p(-\dfrac ml)=l\left(-\dfrac ml\right)+m= -m+m = 0$$.
So, $$x=-\dfrac ml$$ is the zero of the given polynomial $$p(x)$$.  

(vii) $$p(x) =3x^2-1$$:
Put $$x=-\dfrac {1}{\sqrt 3}$$, we get,
$$p(x)=p(-\dfrac{1}{\sqrt3}) = 3\left(-\dfrac {1}{\sqrt 3}\right)^2-1= \left(3 \times \dfrac 13\right)-1 =1-1= 0$$.
So, $$x=-\dfrac {1}{\sqrt3}$$ is the zero of the given polynomial $$p(x)$$.  
Now put $$x=\dfrac {2}{\sqrt 3}$$, we get,
$$p(x) = p(\dfrac{2}{\sqrt3})=3\left(\dfrac {2}{\sqrt 3}\right)^2-1= \left(3 \times \dfrac 43\right)-1 = 4-1=3 \neq 0$$.
So, $$x=\dfrac {2}{\sqrt3}$$ is not the zero of the given polynomial $$p(x)$$.  

(viii) $$p(x) =2x+1$$:
Put $$x=\dfrac 12$$, we get,
$$p(x) = p(\dfrac 12)=2\left(\dfrac 12\right)+1= 1+1 = 2 \neq 0$$.
So, $$x=\dfrac 12$$ is not the zero of the given polynomial $$p(x)$$.  

Mathematics
RS Agarwal
Standard IX

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