Question

if p(x)=x²+4x+7 then find p(0) and p(1)​

Answer 1

Solution :

p(x) = x²-4x+3

i ) p(0) = 0²-4×0+3 = 3

ii ) p(1) = 1²-4×1+3 = 4-4 = 0

iii ) p(2) = 2²-4×2+3 = 7 - 8 = -1

iv ) p(3) = 3²-4×3 + 3 = 12 - 12 = 9

v ) p(x) = x²-4x+3

Let p(x) = 0

=> x²-4x+3 = 0

=> x² - x - 3x + 3 = 0

=> x( x - 1) - 3( x - 1 ) = 0

=> ( x - 1 )( x - 3 ) = 0

=> x -1 = 0 or x - 3 = 0

x = 1 or x = 3

Therefore ,

1 , 3 are zeroes of p(x)

Step-by-step explanation:

please mark me the brainliest

Answer 2

Given: The equation$p(x)=x^{2} +4x+7$.

We have to find$p(0) and p(1)$.

For this, we will put$x=0$ to find$p(0)$ and$x=1$ to find$p(1)$.

We are solving in the following way:

We have,

$p(x)=x^{2} +4x+7$

Now,

We will put$x=0$ to find$p(0)$ and$x=1$ to find$p(1)$.

$=>p(0)=0^{2}+4\times0+7\\=>p(0)=7$

and

$=>p(1)=1^{2}+4\times1+7\\=>p(1)=1+4+7\\=>p(1)=12$

We get, $p(0)=7 and p(1)=12$

Hence, the value of $p(0)=7 and p(1)=12$.