Find the value of the polynomial $ 5 x-4 x^{2}+3 $ at
(i) $ x=0 $
(ii) $ x=-1 $
(iii) $ x=2 $

To do: 

We have to find the value of the polynomial \( 5 x-4 x^{2}+3 \) at

(i) \( x=0 \)
(ii) \( x=-1 \)
(iii) \( x=2 \)

Solution:

To find the value of the polynomial $f(x)$ at $x=a$, we have to substitute $x=a$ in $f(x)$.

Let $f(x)=5 x-4 x^{2}+3$

Therefore,

(i) When $x=0$,

$f(0) = 5 (0)-4 (0)^{2}+3$

$= 0-4 (0)+3$

$= 0-0 + 3$

$= 3$

Hence, the value of the polynomial \( 5 x-4 x^{2}+3 \) at \( x=0 \) is $3$.

(ii) When $x=-1$,

$f(-1) = 5 (-1)-4 (-1)^{2}+3$

$= -5-4 (1)+3$

$= -5-4 + 3$

$= -9+3$

$=-6$

Hence, the value of the polynomial \( 5 x-4 x^{2}+3 \) at \( x=-1 \) is $-6$.

(iii) When $x=2$,

$f(2) = 5 (2)-4 (2)^{2}+3$

$= 10-4 (4)+3$

$= 10-16 + 3$

$= 13-16$

$=-3$

Hence, the value of the polynomial \( 5 x-4 x^{2}+3 \) at \( x=2 \) is $-3$.

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Updated on: 10-Oct-2022

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