
What are the zeroes of the polynomial $x\left( x-2 \right)\left( x+3 \right)$?
Answer
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Hint: In this question, we are given a polynomial $x\left( x-2 \right)\left( x+3 \right)$ whose zeroes we have to find, which means we have to find roots of this polynomial. For this, we equate the polynomial equal to zero, since the polynomial is already factorized so we just put all factors to zero and find different values of x. These values will satisfy the polynomial when equal to zero.
Complete step-by-step solution
Before jumping into finding the solution, let us first understand the meaning of zeroes of polynomials. For any polynomial, there are some values of the variable which makes polynomial zero. These values are called zeroes of the polynomial. These are also called roots of the polynomial. The degree of the polynomial determines the number of roots of the polynomial.
We are given polynomial as $x\left( x-2 \right)\left( x+3 \right)$
Let us first simplify it to find its degree and hence, the number of zeroes it has. Simplifying we get:
$\Rightarrow x\left( {{x}^{2}}-2x+3x-6 \right)={{x}^{3}}+{{x}^{2}}-6x$
Hence, the given polynomial has three roots.
Let us take the original polynomial given which is $x\left( x-2 \right)\left( x+3 \right)$
Since this polynomial is already factorized, so, it will be easy to find zeros by just putting all factors to zero.
Putting all factors to zero one by one we get:
$\Rightarrow x=0$.
$\Rightarrow x-2=0$ simplified as $x = 2$
$\Rightarrow x+3=0$ simplified as $x = -3.$
Hence, we have obtained three different zeroes which are 0, 2, and -3. Putting any of them in a polynomial will give us the value of the polynomial as zero.
Note: Students should try to factorize the polynomial first for finding the zeros of polynomials. Polynomials for which factorization is not possible, we can solve them by various methods depending upon the degree of example. We can use the splitting middle term or discriminant method for a polynomial of degree 2, the hit and trial method can be used for polynomials of degree 3. Students should take care that in the equation $x\left( x-2 \right)\left( x+3 \right)=0$ they should not just take x on the other side and eliminate it. Doing this will give only two roots.
Complete step-by-step solution
Before jumping into finding the solution, let us first understand the meaning of zeroes of polynomials. For any polynomial, there are some values of the variable which makes polynomial zero. These values are called zeroes of the polynomial. These are also called roots of the polynomial. The degree of the polynomial determines the number of roots of the polynomial.
We are given polynomial as $x\left( x-2 \right)\left( x+3 \right)$
Let us first simplify it to find its degree and hence, the number of zeroes it has. Simplifying we get:
$\Rightarrow x\left( {{x}^{2}}-2x+3x-6 \right)={{x}^{3}}+{{x}^{2}}-6x$
Hence, the given polynomial has three roots.
Let us take the original polynomial given which is $x\left( x-2 \right)\left( x+3 \right)$
Since this polynomial is already factorized, so, it will be easy to find zeros by just putting all factors to zero.
Putting all factors to zero one by one we get:
$\Rightarrow x=0$.
$\Rightarrow x-2=0$ simplified as $x = 2$
$\Rightarrow x+3=0$ simplified as $x = -3.$
Hence, we have obtained three different zeroes which are 0, 2, and -3. Putting any of them in a polynomial will give us the value of the polynomial as zero.
Note: Students should try to factorize the polynomial first for finding the zeros of polynomials. Polynomials for which factorization is not possible, we can solve them by various methods depending upon the degree of example. We can use the splitting middle term or discriminant method for a polynomial of degree 2, the hit and trial method can be used for polynomials of degree 3. Students should take care that in the equation $x\left( x-2 \right)\left( x+3 \right)=0$ they should not just take x on the other side and eliminate it. Doing this will give only two roots.
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