Advertisements
Advertisements
प्रश्न
Factorise:
3x3 – x2 – 3x + 1
Advertisements
उत्तर
Let p(x) = 3x3 – x2 – 3x + 1
Constant term of p(x) = 1
Factor of 1 are ±1
By trial, we find that p(1) = 0, so (x – 1) is a factor of p(x)
Now, we see that 3x3 – x2 – 3x + 1
= 3x3 – 3x2 + 2x2 – 2x – x + 1
= 3x2(x – 1) + 2x(x – 1) – 1(x – 1)
= (x – 1)(3x2 + 2x – 1)
Now, (3x2 + 2x – 1) = 3x2 + 3x – x – 1 ...[By splitting middle term]
= 3x(x + 1) – 1(x + 1)
= (x + 1)(3x – 1)
∴ 3x3 – x2 – 3x + 1 = (x – 1)(x + 1)(3x – 1)
APPEARS IN
संबंधित प्रश्न
If `x = 2` is a root of the polynomial `f(x) = 2x2 – 3x + 7a` find the value of a.
Verify whether the indicated numbers is zeros of the polynomials corresponding to them in the following case:
\[p(x) = x^3 - 6 x^2 + 11x - 6, x = 1, 2, 3\]
f(x) = x3 − 6x2 + 2x − 4, g(x) = 1 − 2x
f(x) = x3 −6x2 − 19x + 84, g(x) = x − 7
f(x) = x3 − 6x2 + 11x − 6, g(x) = x2 − 3x + 2
Find the value of a such that (x − 4) is a factors of 5x3 − 7x2 − ax − 28.
Using factor theorem, factorize each of the following polynomials:
x3 + 6x2 + 11x + 6
x3 − 2x2 − x + 2
If x2 − 1 is a factor of ax4 + bx3 + cx2 + dx + e, then
Factorise the following:
(p – q)2 – 6(p – q) – 16
