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
f(x) = 4x4 − 3x3 − 2x2 + x − 7, g(x) = x − 1
If the polynomials ax3 + 3x2 − 13 and 2x3 − 5x + a, when divided by (x − 2) leave the same remainder, find the value of a.
Find the remainder when x3 + 3x2 + 3x + 1 is divided by 5 + 2x .
The polynomials ax3 + 3x2 − 3 and 2x3 − 5x + a when divided by (x − 4) leave the remainders R1 and R2 respectively. Find the value of the following case, if R1 = R2.
In the following two polynomials, find the value of a, if x − a is factor x6 − ax5 + x4 − ax3 + 3x − a + 2.
In the following two polynomials, find the value of a, if x + a is a factor x4 − a2x2 + 3x −a.
If x3 + ax2 − bx+ 10 is divisible by x2 − 3x + 2, find the values of a and b.
x4 − 2x3 − 7x2 + 8x + 12
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:
p² – 6p – 16
