Question

If 0 and 1 are the zeros of the polynomial \[f(x) = 2{x^3} - 3{x^2} + ax + b\], find the value of \[a\] and \[b\].

Answer 1

Hint: In this question, two roots of the polynomial are given. We put those values in the polynomial then the polynomial is equated to \[0\]. From there we get two equations and after that, we have to solve both the equations in order to get the value of \[a\] and \[b\].

Complete step-by-step solution:
Given,
A polynomial \[f(x) = 2{x^3} - 3{x^2} + ax + b\]
And the roots of the polynomial.
\[x = 1\] and
\[x = 0\]
To find,
The value of \[a\] and \[b\].
The polynomial is
\[f(x) = 2{x^3} - 3{x^2} + ax + b\] …………………………….(i)
On putting \[x = 0\] in equation (i) we get,
\[f(0) = 2{(0)^3} - 3{\left( 0 \right)^2} + a\left( 0 \right) + b\]
On further solving
\[f(0) = b\] ……………………………….(ii)
\[x = 0\] is the roots of the polynomial then on putting that value the polynomial must be equal to \[0\].
\[ \Rightarrow f(0) = 0\]
On putting this value in equation (ii) we get,
\[b = 0\]
From here we get the value of b that is
\[b = 0\] …….(iii)
Now on putting \[x = 1\] in the polynomial we get
\[f(1) = 2{(1)^3} - 3{\left( 1 \right)^2} + a\left( 1 \right) + b\]
On further solving
\[f(1) = - 1 + a + b\] ………………………(iv)
\[x = 0\] is the roots of the polynomial then on putting that value the polynomial must be equal to \[0\].
\[ \Rightarrow f(1) = 0\]
On putting this value in equation (iv) we get,
\[0 = - 1 + a + b\]
On putting the value of b from equation (iii)
\[0 = - 1 + a + 0\]
On further solving
\[a = 1\] …….(v)
Final answer:
Hence from equation (iii) and (iv) we get the result
\[ \Rightarrow a = 1\] and
\[ \Rightarrow b = 0\]

Note: If only one root is given in the question and only one variable is given then we put that rooted place of \[x\] and equate that polynomial equal to zero. Then we get the value of that variable according to the equation to which degree the variable exists. If that is linear then we get only one variable and if that is quadratic then we get two values.