Advertisements
Advertisements
प्रश्न
Show that p – 1 is a factor of p10 – 1 and also of p11 – 1.
उत्तर
Let g(p) = p10 – 1 ...(i)
And h(p) = p11 – 1 ...(ii)
On putting p = 1 in equation (i), we get
g(1) = 110 – 1
= 1 – 1
= 0
Hence, p – 1 is a factor of g(p).
Again, putting p = 1 in equation (ii), we get
h(1) = (1)11 – 1
= 1 – 1
= 0
Hence, p – 1 is a factor of h(p).
APPEARS IN
संबंधित प्रश्न
Determine the following polynomial has (x + 1) a factor:
x3 + x2 + x + 1
Factorise:
x3 – 2x2 – x + 2
Determine the following polynomial has (x + 1) a factor:
x4 + 3x3 + 3x2 + x + 1
Factorize the following polynomial.
(x2 – 6x)2 – 8 (x2 – 6x + 8) – 64
The factorisation of 4x2 + 8x + 3 is ______.
Show that 2x – 3 is a factor of x + 2x3 – 9x2 + 12.
Find the value of m so that 2x – 1 be a factor of 8x4 + 4x3 – 16x2 + 10x + m.
Factorise:
84 – 2r – 2r2
Factorise:
`2sqrt(2)a^3 + 8b^3 - 27c^3 + 18sqrt(2)abc`
Without finding the cubes, factorise:
(x – 2y)3 + (2y – 3z)3 + (3z – x)3
