If 2 and 0 are the zeros of the polynomial $f\left( x \right) = 2{x^3} - 5{x^2} + ax + b$ then find the values of ‘a’ and ‘b’?
Answer
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Hint: Here, we are asked to solve the given polynomial to find the values of ‘a’ and ‘b’. From given the real numbers, m and n are zeros of polynomial $f\left( x \right)$ so that $f\left( m \right)$$ = 0$ and $f\left( n \right) = 0$
Now here are the some of the types of polynomials that you will come across while studying them:
$\bullet$ Value of Polynomial and Division Algorithm, Degree of Polynomial, Factorization of Polynomials,
$\bullet$ Remainder Theorem, Factor Theorem, and Geometrical Representation of Zeroes of a Polynomial.
Complete step-by-step solution:
Given that $2$ and $0$ are the zeros of the polynomial $f\left( x \right) = 2{x^3} - 5{x^2} + ax + b$,
So that, $f\left( 2 \right) = 0$ and $f\left( 0 \right) = 0$
First, we taking $x = 2$
Putting the value $x = 2$ in the given polynomial,
$f\left( 2 \right) = 2{\left( 2 \right)^3} - 5{\left( x \right)^2} + 2a + b$
$ = 2\left( 8 \right) - 5\left( 4 \right) + 2a + b$
$ = 16 - 20 + 2a + b$
$ = - 4 + 2a + b$
Since we know, $f\left( 2 \right) = 0$
$ - 4 + 2a + b = 0$
Therefore, $2a + b = 4$ ………………………………… {1}
Now, we taking $x = 0$
Putting the value $x = 0$in the given polynomial,
$f\left( 0 \right) = 2{\left( 0 \right)^3} - 5{\left( 0 \right)^2} + 0a + b$
$ = 0 - 0 + 0 + b$
$ = b$
Since we know, $f\left( 0 \right) = 0$
Therefore, $b = 0$ ………………………………. {2}
Now, we putting the value of b {2} in equation {1},
$2a + b = 4$
$\Rightarrow 2a + 0 = 4$
$\Rightarrow 2a = 4$
$\Rightarrow a = \dfrac{4}{2}$
$\Rightarrow a = 2$
Hence, the value of ‘a’ and ‘b’ are $2$ and $0$.
Note: We already know that a polynomial is an algebraic term with one or many terms. The zeros of a polynomial are the real values of the variables for which the value of the polynomial becomes zero. So that the real numbers, ‘m’ and ‘n’ are zeros of the polynomial p(x), if p(m)=0 and p(n)=0. Sometimes the zeros of polynomials are also called as Roots of the polynomials. The standard form of a polynomial in x is ${a_n}{x^n} + {a_{n - 1}}{x^{n - 1}} + ...... + {a_1}x + {a_0}$, where ${a_n},{a_{n - 1}},.....,{a_1},{a_0}$ are constants, ${a_n} \ne 0$ and n is a whole number.
By now you are aware of the zero polynomial.
Now here are the some of the types of polynomials that you will come across while studying them:
$\bullet$ Value of Polynomial and Division Algorithm, Degree of Polynomial, Factorization of Polynomials,
$\bullet$ Remainder Theorem, Factor Theorem, and Geometrical Representation of Zeroes of a Polynomial.
Complete step-by-step solution:
Given that $2$ and $0$ are the zeros of the polynomial $f\left( x \right) = 2{x^3} - 5{x^2} + ax + b$,
So that, $f\left( 2 \right) = 0$ and $f\left( 0 \right) = 0$
First, we taking $x = 2$
Putting the value $x = 2$ in the given polynomial,
$f\left( 2 \right) = 2{\left( 2 \right)^3} - 5{\left( x \right)^2} + 2a + b$
$ = 2\left( 8 \right) - 5\left( 4 \right) + 2a + b$
$ = 16 - 20 + 2a + b$
$ = - 4 + 2a + b$
Since we know, $f\left( 2 \right) = 0$
$ - 4 + 2a + b = 0$
Therefore, $2a + b = 4$ ………………………………… {1}
Now, we taking $x = 0$
Putting the value $x = 0$in the given polynomial,
$f\left( 0 \right) = 2{\left( 0 \right)^3} - 5{\left( 0 \right)^2} + 0a + b$
$ = 0 - 0 + 0 + b$
$ = b$
Since we know, $f\left( 0 \right) = 0$
Therefore, $b = 0$ ………………………………. {2}
Now, we putting the value of b {2} in equation {1},
$2a + b = 4$
$\Rightarrow 2a + 0 = 4$
$\Rightarrow 2a = 4$
$\Rightarrow a = \dfrac{4}{2}$
$\Rightarrow a = 2$
Hence, the value of ‘a’ and ‘b’ are $2$ and $0$.
Note: We already know that a polynomial is an algebraic term with one or many terms. The zeros of a polynomial are the real values of the variables for which the value of the polynomial becomes zero. So that the real numbers, ‘m’ and ‘n’ are zeros of the polynomial p(x), if p(m)=0 and p(n)=0. Sometimes the zeros of polynomials are also called as Roots of the polynomials. The standard form of a polynomial in x is ${a_n}{x^n} + {a_{n - 1}}{x^{n - 1}} + ...... + {a_1}x + {a_0}$, where ${a_n},{a_{n - 1}},.....,{a_1},{a_0}$ are constants, ${a_n} \ne 0$ and n is a whole number.
By now you are aware of the zero polynomial.
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