Advertisements
Advertisements
प्रश्न
Show that x + 3 is a factor of 69 + 11x – x2 + x3.
Advertisements
उत्तर
Let p(x) = x3 – x2 + 11x + 69
We have to show that, x + 3 is a factor of p(x).
i.e., p(–3) = 0
Now, p(–3) = (–3)3 – (–3)2 + 11(–3) + 69
= –27 – 9 – 33 + 69
= – 69 + 69
= 0
Hence, (x + 3) is a factor of p(x).
APPEARS IN
संबंधित प्रश्न
Use the Factor Theorem to determine whether g(x) is a factor of p(x) in the following case:
p(x) = x3 + 3x2 + 3x + 1, g(x) = x + 2
Find the value of k, if x – 1 is a factor of p(x) in the following case:
p(x) = `kx^2 - sqrt2x +1`
Find the value of k, if x – 1 is a factor of p(x) in the following case:
p(x) = kx2 – 3x + k
Factorise:
6x2 + 5x – 6
Find the Factors of the Polynomial Given Below.
2x2 + x – 1
Factorize the following polynomial.
(y2 + 5y) (y2 + 5y – 2) – 24
Determine which of the following polynomials has x – 2 a factor:
3x2 + 6x – 24
Show that p – 1 is a factor of p10 – 1 and also of p11 – 1.
If x + 1 is a factor of ax3 + x2 – 2x + 4a – 9, find the value of a.
Without finding the cubes, factorise:
(x – 2y)3 + (2y – 3z)3 + (3z – x)3
