Advertisements
Advertisements
प्रश्न
Factorise:
2x3 – 3x2 – 17x + 30
Advertisements
उत्तर
Let p(x) = 2x3 – 3x2 – 17x + 30
Constant term of p(x) = 30
∴ Factors of 30 are ±1, ±2, ±3, ±5, ±6, ±10, ±15, ±30
By trial, we find that p(2) = 0, so (x – 2) is a factor of p(x) ...[∵ 2(2)3 – 3(2)2 – 17(2) + 30 = 16 – 12 – 34 + 30 = 0]
Now, we see that 2x3 – 3x2 – 17x + 30
= 2x3 – 4x2 + x2 – 2x – 15x + 30
= 2x2(x – 2) + x(x – 2) – 15(x – 2)
= (x – 2)(2x2 + x – 15) ...[Taking (x – 2) common factor]
Now, (2x2 + x – 15) can be factorised either by splitting the middle term or by using the factor theorem.
Now, (2x2 – x – 15) = 2x2 + 6x – 5x – 15 ...[By splitting the middle term]
= 2x(x + 3) – 5(x + 3)
= (x + 3)(2x – 5)
∴ 2x3 – 3x2 – 17x + 30 = (x – 2)(x + 3)(2x – 5)
APPEARS IN
संबंधित प्रश्न
Write the coefficient of x2 in the following:
`17 -2x + 7x^2`
Write the coefficient of x2 in the following:
`9-12x +X^3`
Verify whether the indicated numbers is zeros of the polynomials corresponding to them in the following case:
\[p(x) = x^3 - 6 x^2 + 11x - 6, x = 1, 2, 3\]
Find rational roots of the polynomial f(x) = 2x3 + x2 − 7x − 6.
In the following two polynomials, find the value of a, if x + a is a factor x3 + ax2 − 2x +a + 4.
Find the values of p and q so that x4 + px3 + 2x3 − 3x + q is divisible by (x2 − 1).
y3 − 2y2 − 29y − 42
If x3 + 6x2 + 4x + k is exactly divisible by x + 2, then k =
Factorise the following:
`1/x^2 + 1/y^2 + 2/(xy)`
If (x + 5) and (x – 3) are the factors of ax2 + bx + c, then values of a, b and c are
