Question

if p(x) = x square - 4x + 3. Evaluate p(2)-pc(-1) + p(1/2)​

Answer 1

Answer:

After putting x=2 in equation, we get −1.

After putting x=−1, we get 8

After putting x=1/2 we get 5/4

So, p(2)−p(−1)+p(1/2) = -1 -8 + 5/4

=−9+ 5/4 = -36+ 5/4 = -31/4

Answer 2

Appropriate Question :-

$\sf \: If \: p(x) = {x}^{2} - 4x + 3, \: evaluate \: p(2) - p( - 1) + p\bigg(\dfrac{1}{2} \bigg)$

$\large\underline{\sf{Solution-}}$

Given polynomial is

$\rm \: p(x) = {x}^{2} - 4x + 3$

Consider

$\rm \: p(2)$

$\rm \: =  \: {2}^{2} - 4(2) + 3$

$\rm \: =  \: 4 - 8 + 3$

$\rm \: =  \: 7 - 8$

$\rm \: =  \: - 1$

$\rm\implies \:\boxed{\sf{  \:\rm \: p(2) \: = \: - \: 1 \: \:}}$

Now, Consider

$\rm \: p( - 1)$

$\rm \: =  \: {( - 1)}^{2} - 4( - 1) + 3$

$\rm \: =  \: 1 + 4 + 3$

$\rm \: =  \: 8$

$\rm\implies \:\boxed{\sf{  \:\rm \: p( - 1) \: = \: 8 \: \:}}$

Now, Consider

$\rm \: p\bigg(\dfrac{1}{2} \bigg)$

$\rm \: =  \: {\bigg(\dfrac{1}{2} \bigg) }^{2} - 4 \times \dfrac{1}{2} + 3$

$\rm \: =  \:\dfrac{1}{4} - 2 + 3$

$\rm \: =  \:\dfrac{1}{4} + 1$

$\rm \: =  \:\dfrac{1 + 4}{4}$

$\rm \: =  \:\dfrac{5}{4}$

$\rm\implies \:\boxed{\sf{  \:\rm \: p\bigg(\dfrac{1}{2} \bigg) = \dfrac{5}{4} \: \: }}$

Now, Consider

$\rm \: p(2) - p( - 1) + p\bigg(\dfrac{1}{2} \bigg)$

$\rm \: =  \: - 1 - 8 + \dfrac{5}{4}$

$\rm \: =  \: - 9 + \dfrac{5}{4}$

$\rm \: =  \: \dfrac{ - 36 + 5}{4}$

$\rm \: =  \: - \: \dfrac{31}{4}$

Hence,

$\rm\implies \:\boxed{\sf{  \:\rm \: p(2) - p( - 1) + p\bigg(\dfrac{1}{2} \bigg) = - \frac{31}{4} \: \: }}$

$\rule{190pt}{2pt}$

Additional Information :-

Algebraic Identities to know : -

➢  (a + b)² = a² + 2ab + b²

➢  (a - b)² = a² - 2ab + b²

➢  a² - b² = (a + b)(a - b)

➢  (a + b)² = (a - b)² + 4ab

➢  (a - b)² = (a + b)² - 4ab

➢  (a + b)² + (a - b)² = 2(a² + b²)

➢  (a + b)³ = a³ + b³ + 3ab(a + b)

➢  (a - b)³ = a³ - b³ - 3ab(a - b)