Question

The polynomials p(t) = 4t3 - st2 + 7 and q(t) = t2 + st + 8 leave the same remainder when divided by (t - 1). find the value of s.

Answer 1

According to the question answer is
s=1

Answer 2

Answer:

s = 1

Step-by-step explanation:

Remainder theorem: If a polynomial P(x) is divided by (x-c), then the remainder is equal to P(c).

The given polynomial are

$p(t)=4t^3-st^2+7$

$q(t)=t^2+st+8$

Using remainder theorem the remainder of $\frac{p(t)}{t-1}$ is p(1) and the remainder of $\frac{q(t)}{t-1}$ is q(1).

Substitute t=1 in the given functions.

$p(1)=4(1)^3-s(1)^2+7\Rightarrow 4-s+7=11-s$

$q(1)=(1)^2+s(1)+8=1+s+8=9+s$

It is given that if p(t) and q(t) divided by (t-1), then the remainder is same.

$p(1)=q(1)$

Substitute these values.

$11-s=9+s$

Add s on both sides.

$11=9+s+s$

$11=9+2s$

Subtract 9 from both sides.

$11-9=2s$

$2=2s$

Divide both sides by 2.

$1=s$

Therefore, the value of s is 1.