Advertisements
Advertisements
प्रश्न
In the expansion of (2x2 - 8) (x - 4)2; find the value of constant term.
Advertisements
उत्तर
( 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
APPEARS IN
संबंधित प्रश्न
Expand: `( 2x - 1/x )( 3x + 2/x )`
Expand : ( 5x - 3y - 2 )2
Expand : `( x - 1/x + 5)^2`
If a + b + c = 12 and a2 + b2 + c2 = 50; find ab + bc + ca.
If a + `1/a` = m and a ≠ 0 ; find in terms of 'm'; the value of :
`a - 1/a`
In the expansion of (2x2 - 8) (x - 4)2; find the value of coefficient of x3.
If x > 0 and `x^2 + 1/[9x^2] = 25/36, "Find" x^3 + 1/[27x^3]`
The difference between two positive numbers is 4 and the difference between their cubes is 316.
Find : Their product
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`
