Question

find p(0),p(1)and p(2) for
2x²+7x-1​

Answer 1

Answer:

p(0)=2(0)^2+7(0)-1

=0+0-1

=-1

p(1)=2(1)^2+7(1)-1

= 2+7-1

=9-1

=8

Step-by-step explanation:

p(2)=2(2)^2+7(2)-1

=2(4)+14-1

=8+14-1

=22-1

=21

Answer 2

Answer:

Question :-

find p(0),p(1)and p(2) for

Required Answer :-

p(0) = -1

p(1) = 8

p(2) = 21

Solution :-

let's first find p(0)

for which we have to substitute 0 in place of of x

${2x}^{2} + 7x - 1$

$= 2 \times 0 \times 0 + 7 \times 0 - 1$

$= - 1$

Now let's find for p(1)

${2x}^{2} + 7x - 1$

$2 \times 1 \times 1 + 7 \times 1 - 1$

$= 2 + 7 - 1$

$= 8$

Now p(2)

${2x}^{2} + 7x - 1$

$2 \times 2 \times 2 + 7 \times 2 - 1$

$8 + 14 - 1$

$= 21$

Therefore p(0) = -1

p(1) = 8

p(2) = 21