Question

If the polynomials ty3+4y2+3y−4 and y3−4y+t leave the same remainder when divided by (y−3), find the value of t.

Answer 1

Required Answer :

★ The value of t = 5/13

Solution :

Let the remainder in both the cases be R.

→ f(x) = ty³ + 4y² + 3y - 4 = R

→ P(x) = y³ - 4y + t = R

Both the polynomials are divided by (y - 3).

→ y - 3 = 0

→ y = 3

Substituting the value of y in the first polynomial :

→ f(x) = ty³ + 4y² + 3y - 4 = R

→ t(3)³ + 3(3) - 4 = R

→ 27t + 9 - 4 = R

→ 27t + 5 = R ----(1)

Substituting the value of y in the second polynomial :

→ P(x) = y³ - 4y + t = R

→ (3)³ - 4(3) + t = R

→ 27 - 12 + t = R

→ 15 + t = R ----(2)

Comparing equation (1) and (2) :

→ 27t + 5 = 15 + t

→ 27t - t = 15 - 5

→ 26t = 10

→ t = 10/26

→ t = 5/13

Therefore, the value of t = 5/13