Question

if p(x) =2x^2-7x+1, then evaluate p(3)+p(1/2)+p(-2) ​

Answer 1

Step-by-step explanation:

p(x)=2x²-7x-1

p(3)=2(3)²-7(3)-1

=2(9)-21-1

18-22=-4. ......................(1)

p(1/2)=2(1/2)²-7(1/2)-1

=1/2-7/2-1

-6/2-1

-6-2/2=-8/2=-4...............(2)

p(-2)=2(-2)²-7(-2)+1

=2(4)+14+1

8+15=23...........(3)

adding both eqn (1)(2)(3)

p(3)+p(1/2)+p(-2)

-4-4+23

-8+23

=15 OK friend this is urs answer

Answer 2

Answer:

$\\p=\frac{2x^2-7x+1}{x}$

$=\frac{6x^3-21x^2+3}{2x}$

Step-by-step explanation:

Let's solve for p.

px=2x2−7x+1

Step 1: Divide both sides by x.

$\frac{px}{x}=\frac{2x^2-7x+1}{x} \\p=\frac{2x^2-7x+1}{x}$

Evaluate for p=2x2−7x+$\frac{1}{x}$

(2x2−7x+$\frac{1}{x}$) (3)+(2x2−7x+$\frac{1}{x}$) $(\frac{1}{2} )+(2x^2-7x+\frac{1}{x} )(-2)$

answer:-,$=\frac{6x^3-21x^2+3}{2x}$