Question

find p(1),p(-2) and p(1/2) of
1.p(z) = 2 z^2 -3 z + 4 2. p(t) = 3 t^3 - 2 t^2 + t

Answer 1

1) p(z)=2z^2-3z+4
so,
p(1)=2(1)^2-3+4
=2-3+4
=3

p(-2)=2(-2)^2-3(-2)+4
=8+6+4
=18

p(1/2)=2(1/2)^2-3/2+4
=1/2-3/2+4
=(-2/2)+4
=(-1+4)
=3

Similarly,p(t)=3t^3-2t^2+t

p(1)=3(1)^3-2(1)^2+1
=3-2+1
=2

p(-2)=3(-2)^3-2(-2)^2+(-2)
=3(-8)-2(4)-2
=(-24 -8-2)
=(-34)

p(1/2)=3(1/2)^3-2(1/2)^2 +1/2
=3/8 - 1/2+1/2
=3/8

PLZ MARK IT AS BRAINLIEST

Answer 2

`Hey there !!`

p(z)=2z^2-3z+4

p(1) = 2×1²-3+4
=2-3+4
=3
------------------------------------

p(-2) = 2×[-2]²-3 × -2+4
= 8+6+4
=18
--------------------------------------------

p(1/2) =2 × [1/2]²-3/2+4
=1/2 -3/2+4 =
=(-1+4)
=3
=======================================================================

p(t)=3t³-2t²+t

p(1) = 3 × 1³-2 × 1²+1
= 3-2+1
= 2
---------------------------

p(-2) = 3 ×(-2)³-2(-2)²+(-2)
= 3 × (-8) -2 × 4 -2
= -24 - 8 - 2
= -34
-------------------------------------

p(1/2) =3(1/2)³-2(1/2)² +1/2
= 3/8 - 1/2+1/2
= 3/8