Advertisements
Advertisements
प्रश्न
Show that p – 1 is a factor of p10 – 1 and also of p11 – 1.
Advertisements
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
Find the value of k, if x – 1 is a factor of p(x) in the following case:
p(x) = kx2 – 3x + k
Factorize the following polynomial.
(y + 2) (y – 3) (y + 8) (y + 3) + 56
One of the factors of (25x2 – 1) + (1 + 5x)2 is ______.
Show that x + 3 is a factor of 69 + 11x – x2 + x3.
Determine which of the following polynomials has x – 2 a factor:
3x2 + 6x – 24
Factorise:
84 – 2r – 2r2
Factorise the following:
9x2 – 12x + 4
Factorise:
a3 – 8b3 – 64c3 – 24abc
Factorise:
`2sqrt(2)a^3 + 8b^3 - 27c^3 + 18sqrt(2)abc`
If both x – 2 and `x - 1/2` are factors of px2 + 5x + r, show that p = r.
