Advertisements
Advertisements
Question
Find the value of k, if x – 1 is a factor of p(x) in the following case:
p(x) = `2x^2+kx+sqrt2`
Advertisements
Solution
If x − 1 is a factor of polynomial p(x), then p(1) must be 0.
p(x) = `2x^2+kx+sqrt2`
p(1) = 0
⇒ `2(1)^2 + k(1) + sqrt2 = 0`
⇒ `2 + k + sqrt2 = 0`
⇒ k = `-2 -sqrt2`
k = `-(2+sqrt2)`
Therefore, the value of k is `-(2+sqrt2)`.
APPEARS IN
RELATED QUESTIONS
Factorise:
6x2 + 5x – 6
Factorize the following polynomial.
(x2 – 6x)2 – 8 (x2 – 6x + 8) – 64
x + 1 is a factor of the polynomial ______.
One of the factors of (25x2 – 1) + (1 + 5x)2 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.
If x + 1 is a factor of ax3 + x2 – 2x + 4a – 9, find the value of a.
Factorise:
84 – 2r – 2r2
Factorise the following:
1 – 64a3 – 12a + 48a2
Factorise:
`2sqrt(2)a^3 + 8b^3 - 27c^3 + 18sqrt(2)abc`
