Question

If p(t)=t^3-3t^2+t-4 then find p(1) and p(-1)

Answer 1

p(t) = t³ - 3t² + t-4

p(1) = 1³ - 3(1²) + 1-4

= 1 - 3 + (-3)

= -2 -3

= -5

p(-1) = -1³ - 3(-1²) + (-1 -4)

= -1 - 3 + (-5)

= -4 - 5

= -9

Answer 2

Given,

p(t) = $t {}^{3} - 3t {}^{2} + t - 4$

To find,

The value of p(1) and p(-1).

Solution,

The value of p(1) and p(-1) will be -5 and -9 respectively.

According to the question,

$t {}^{3} - 3t {}^{2} + t - 4$

Now, putting 1 in place of t to find the value of p(1):

p(1) = (1)³-3(1)²+1-4

p(1) = 1-3+1-4

p(1) = 2-7

p(1) = -5

Now, putting -1 in place of t to find the value of p(-1):

p(-1) = (-1)³-3(-1)²+(-1)-4

p(-1) = -1-3(1)-1-4

p(-1) = -1-3-1-4

p(-1) = -9

(Note that the two negative integers are added but their result is negative.)

Hence, the value of p(1) and p(-1) are -5 and -9 respectively.