In the Expansion of (2x2 - 8) (X - 4)2; Find the Value of Constant Term.

In the expansion of (2x² - 8) (x - 4)²; find the value of constant term.

Sum

( 2x² - 8 )( x - 4 )²
= ( 2x² - 8 )( x² - 8x + 16 )
= 4x⁴ - 16x³ + 32x² - 8x² + 64x -128
= 4x⁴ - 16x³ + 24x² + 64x - 128
Hence,
Constant term = -128

shaalaa.com

Expansion of Formula

Is there an error in this question or solution?

Chapter 4: Expansions - Exercise 4 (D) [Page 64]

Chapter 4 Expansions
Exercise 4 (D) | Q 3.3 | Page 64

Expand : ( x + 8 ) ( x + 10 )

Expand : ( x - 8 )( x - 10 )

Expand : ( 5a - 3b + c )²

If a² + b² + c² = 35 and ab + bc + ca = 23; find a + b + c.

If x + 2y + 3z = 0 and x³ + 4y³ + 9z³ = 18xyz ; evaluate :
`[( x + 2y )^2]/(xy) + [(2y + 3z)^2]/(yz) + [(3z + x)^2]/(zx)`

If a + `1/a` = m and a ≠ 0 ; find in terms of 'm'; the value of :
`a - 1/a`

If 2( x² + 1 ) = 5x, find :
(i) `x - 1/x`

(ii) `x^3 - 1/x^3`

If `(x^2 + 1)/x = 3 1/3` and x > 1; find If `x^3 - 1/x^3`

If 3a + 5b + 4c = 0, show that : 27a³ + 125b³ + 64c³ = 180 abc

If x = `1/[ 5 - x ] "and x ≠ 5 find "x^3 + 1/x^3`