If the polynomial ax³ + bx² + x – 6 is divided by x – 2 and remainder is 4, then let us calculate the values of a and b when x + 2 is a factor of this polynomial.

Formula used.

If f(x) is a polynomial with degree n

Then (x – a) is a factor of f(x) if f(a) = 0

⇒ Dividend = Divisor × Quotient + Remainder

When x + 2 is factor of polynomial

Then

x + 2 = 0

x = – 2

if x + 2 is factor of polynomial f(x) = ax³ + bx² + x – 6

then f( – 2) is 0

f( – 2) = a( – 2)³ + b( – 2)² + ( – 2) – 6 = 0

– 8a + 4b – 2 – 6 = 0

4b – 8a = 8

4b = 8 + 8a ……… eq 1

When x – 2 divides polynomial gives remainder 4

Then

Dividend = Divisor × Quotient + Remainder

ax³ + bx² + x – 6 = (x – 2) × Quotient + 4

ax³ + bx² + x – 6 – 4 = (x – 2) × Quotient

ax³ + bx² + x – 10 = (x – 2) × Quotient

(x – 2) = 0

x = 2

if (x – 2) is factor of polynomial ax³ + bx² + x – 10

then f(2) = 0

f(2) = a(2)³ + b(2)² + 2 – 10

8a + 4b – 8 = 0

8a + 4b = 8 ………eq 2

Putting value of 4b from eq 1 into eq 2

8a + (8 + 8a) = 8

16a + 8 = 8

16a = 8 – 8 = 0

a = 0

Putting value of ‘a’ in eq 1

4b = 8 + 8a

4b = 8 + 8 × 0

4b = 8

b = = 2

Conclusion.

∴ The value of a and b comes to be 0 and 2 respectively