Advertisements
Advertisements
प्रश्न
Factorise:
x3 – 6x2 + 11x – 6
Advertisements
उत्तर
Let p(x) = x3 – 6x2 + 11x – 6
Constant term of p(x) = – 6
Factors of – 6 are ±1, ±2, ±3, ±6.
By trial, we find that p(1) = 0, so (x – 1) is a factor of p(x) ...[∵ (1)3 – 6(1)2 + 11(1) – 6 = 1 – 6 + 11 – 6 = 0]
Now, we see that x3 – 6x2 + 11x = 6
= x3 – x2 – 5x2 + 5x + 6x – 6
= x2(x – 1) – 5x(x – 1) + 6(x – 1)
= (x – 1)(x2 – 5x + 6) ...[Taking (x – 1) common factor]
Now, (x2 – 5x + 6) = x2 – 3x – 2x + 6 ...[By splitting the middle term]
= x(x – 3) – 2(x – 2)
= (x – 3)(x – 2)
∴ x3 – 6x2 + 11x – 6 = (x – 1)(x – 2)(x – 3)
APPEARS IN
संबंधित प्रश्न
If f(x) = 2x2 - 13x2 + 17x + 12 find f(-3).
f(x) = 3x4 + 17x3 + 9x2 − 7x − 10; g(x) = x + 5
In the following two polynomials, find the value of a, if x + a is a factor x3 + ax2 − 2x +a + 4.
What must be added to x3 − 3x2 − 12x + 19 so that the result is exactly divisibly by x2 + x - 6 ?
If x + 2 is a factor of x2 + mx + 14, then m =
The value of k for which x − 1 is a factor of 4x3 + 3x2 − 4x + k, is
If x + 2 and x − 1 are the factors of x3 + 10x2 + mx + n, then the values of m and n are respectively
Let f(x) be a polynomial such that \[f\left( - \frac{1}{2} \right)\] = 0, then a factor of f(x) is
Factorise the following:
9 – 18x + 8x2
Factorise the following:
8m3 – 2m2n – 15mn2
