Question

If a polynomial is represented by p(x)=
X^2+ 7x + 8, the value of p(-1)is

Answer 1

AnsWer :

2.

SolutioN :

Case.

$Condition\begin{cases} \sf{p(x) ={x}^{2} + 7x + 8. }\\ \sf{p( - 1) = ?}\end{cases}$

Let's Find the P( - 1 ).

★ We have, Expression.

$\tt \longmapsto {x}^{2} + 7x + 8.$

When,

• P ( - 1 )

$\tt \longmapsto {( - 1)} + 7( - 1) + 8.$

$\tt \longmapsto 1 - 7 + 8.$

$\tt \longmapsto 9 - 7.$

$\tt \longmapsto 2.$

So, When We putting the Value of P ( x ) Where x is - 1 then our Expression become 2.

Therefore, the required answer is 2.

Answer 2

✯✯ QUESTION ✯✯

If a polynomial is represented by p(x)=

x²+ 7x + 8, the value of p(-1) is: -

━━━━━━━━━━━━━━━━━━━━

✰✰ ANSWER ✰✰

$\longmapsto\tt{p(x)={x}^{2}+7x+8}$

$\longmapsto\tt{p(-1)={x}^{2}+7x+8}$

➥Putting x = -1 : -

$\longmapsto\tt{{(-1)}^{2}+7(-1)+8}$

$\longmapsto\tt{1-7+8}$

$\longmapsto\tt{-6+8}$

$\longmapsto\tt{\large{\boxed{\bold{\bold{\green{\sf{2}}}}}}}$

☛Answer is 2....