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:
x3 + 13x2 + 32x + 20
Determine the following polynomial has (x + 1) a factor:
`x^3-x^2-(2+sqrt2)x+sqrt2`
Find the Factors of the Polynomial Given Below.
2x2 + x – 1
Factorize the following polynomial.
(x2 – x)2 – 8 (x2 – x) + 12
x + 1 is a factor of the polynomial ______.
Show that 2x – 3 is a factor of x + 2x3 – 9x2 + 12.
Determine which of the following polynomials has x – 2 a factor:
3x2 + 6x – 24
Factorise:
84 – 2r – 2r2
Factorise the following:
1 – 64a3 – 12a + 48a2
If both x – 2 and `x - 1/2` are factors of px2 + 5x + r, show that p = r.
