
Find the zero of the polynomial in each of the following cases:
(i)$p(x) = x + 5$
(ii)$p(x) = x - 5$
(iii)$p(x) = 2x + 5$
(iv)$p(x) = 3x - 2$
(v)$p(x) = 3x$
(vi)$p(x) = ax,a \ne 0$
(vii)$p(x) = cx + d,c \ne 0,c,d$ are real numbers.
Answer
576.3k+ views
Hint: A number $x$ is called the zero of a polynomial if $p(x) = 0$. So to find the zero of the polynomials, equate the given expressions to zero and then evaluate the value of $x$ from it. Use this concept to calculate the zero of the given polynomials.
Complete step-by-step answer:
To find the zero of the polynomials, we will set each $p(x) = 0$ and then evaluate for $x$.
(i):
Setting $p(x) = 0$ , we get,
$
x + 5 = 0 \\
\Rightarrow x = - 5 \\
$
Therefore, the zero of the polynomial is $x = - 5$.
(ii):
Again, setting $p(x) = 0$ , we get,
$
x - 5 = 0 \\
\Rightarrow x = 5 \\
$
Therefore, the zero of the polynomial is $x = 5$.
(iii):
We set $p(x) = 0$to get,
$
2x + 5 = 0 \\
\Rightarrow 2x = - 5 \\
\Rightarrow x = - \dfrac{5}{2} \\
$
Therefore, the zero of the polynomial is $x = - \dfrac{5}{2}$ .
(iv):
Like before, we set $p(x) = 0$ to obtain,
$
3x - 2 = 0 \\
\Rightarrow 3x = 2 \\
\Rightarrow x = \dfrac{2}{3} \\
$
Therefore, the zero of the polynomial is $x = \dfrac{2}{3}$.
(v):
We again set $p(x) = 0$ to obtain,
$
3x = 0 \\
\Rightarrow x = 0 \\
$
Therefore, the zero of the polynomial is $x = 0$.
(vi):
Setting $p(x) = 0$ , we get,
$ax = 0$
So either $a$ or $x$ has to be $0$.
But since $a \ne 0$ , $x = 0$.
Therefore, the zero of the polynomial is $x = 0$.
(vii):
Setting $p(x) = 0$ , we get,
$
cx + d = 0 \\
\Rightarrow cx = - d \\
\Rightarrow x = - \dfrac{d}{c} \\
$
Therefore the zero of this polynomial is $x = - \dfrac{d}{c}$ .
Note: The significance of mentioning $a \ne 0$ and $c \ne 0$ in the question is that if they had been zero, the value of $x$ would have been a fraction whose denominator would be zero. That would make $x$ either invalid or infinity depending on the value of the numerator. So to sum up, to find the zero of a polynomial, equate the expression with zero and then evaluate the value of $x$. Other than this, make sure that you do not skip steps as that can lead to calculation mistakes.
Complete step-by-step answer:
To find the zero of the polynomials, we will set each $p(x) = 0$ and then evaluate for $x$.
(i):
Setting $p(x) = 0$ , we get,
$
x + 5 = 0 \\
\Rightarrow x = - 5 \\
$
Therefore, the zero of the polynomial is $x = - 5$.
(ii):
Again, setting $p(x) = 0$ , we get,
$
x - 5 = 0 \\
\Rightarrow x = 5 \\
$
Therefore, the zero of the polynomial is $x = 5$.
(iii):
We set $p(x) = 0$to get,
$
2x + 5 = 0 \\
\Rightarrow 2x = - 5 \\
\Rightarrow x = - \dfrac{5}{2} \\
$
Therefore, the zero of the polynomial is $x = - \dfrac{5}{2}$ .
(iv):
Like before, we set $p(x) = 0$ to obtain,
$
3x - 2 = 0 \\
\Rightarrow 3x = 2 \\
\Rightarrow x = \dfrac{2}{3} \\
$
Therefore, the zero of the polynomial is $x = \dfrac{2}{3}$.
(v):
We again set $p(x) = 0$ to obtain,
$
3x = 0 \\
\Rightarrow x = 0 \\
$
Therefore, the zero of the polynomial is $x = 0$.
(vi):
Setting $p(x) = 0$ , we get,
$ax = 0$
So either $a$ or $x$ has to be $0$.
But since $a \ne 0$ , $x = 0$.
Therefore, the zero of the polynomial is $x = 0$.
(vii):
Setting $p(x) = 0$ , we get,
$
cx + d = 0 \\
\Rightarrow cx = - d \\
\Rightarrow x = - \dfrac{d}{c} \\
$
Therefore the zero of this polynomial is $x = - \dfrac{d}{c}$ .
Note: The significance of mentioning $a \ne 0$ and $c \ne 0$ in the question is that if they had been zero, the value of $x$ would have been a fraction whose denominator would be zero. That would make $x$ either invalid or infinity depending on the value of the numerator. So to sum up, to find the zero of a polynomial, equate the expression with zero and then evaluate the value of $x$. Other than this, make sure that you do not skip steps as that can lead to calculation mistakes.
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