Question

The coefficient of x³ in the expansion of (3 +bx )⁵ is −720. Find the value of the constant b.​

Answer 1

Answer:

b = -2

Step-by-step explanation:

We need to use the binomial theorem, to expand

the expression (3 + bx )⁵ :

(3 + bx )⁵ = 3⁵ + 5(3)⁴(bx)¹ + 10(3)³(bx)² + 10(3)²(bx)³ + 5(3)¹(bx)⁴ + (bx)⁵

             

               = 243 + 5×81×(bx)¹ + 10×27×(bx)² + 10×9×(bx)³ + 5×3×(bx)⁴ + (bx)⁵

               = 243 + 405(bx)¹ + 270(bx)² + 90(bx)³ + 15(bx)⁴ + (bx)⁵

               = 243 + (405b)x + (270b²)x² + (90b³)x³ + (15b⁴)x⁴ + (bx)⁵

Then

(3 + bx )⁵ = 243 + (405b)x + (270b²)x² + (90b³)x³ + (15b⁴)x⁴ + (bx)⁵

Then

The coefficient of x³ in the expansion of (3 + bx )⁵ is 90b³

Comparing the coefficients :

90b³ = -720

Then

b³ = (-720) ÷ 90 = -8

Then

b³ = -8

Then

b = -2