Advertisements
Advertisements
Question
Factorise:
2x3 – 3x2 – 17x + 30
Advertisements
Solution
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
RELATED QUESTIONS
Identify constant, linear, quadratic and cubic polynomials from the following polynomials:
`h(x)=-3x+1/2`
If `x = 2` is a root of the polynomial `f(x) = 2x2 – 3x + 7a` find the value of a.
\[f(x) = 3 x^4 + 2 x^3 - \frac{x^2}{3} - \frac{x}{9} + \frac{2}{27}, g(x) = x + \frac{2}{3}\]
In each of the following, use factor theorem to find whether polynomial g(x) is a factor of polynomial f(x) or, not: (1−7)
f(x) = x3 − 6x2 + 11x − 6; g(x) = x − 3
Find the values of a and b so that (x + 1) and (x − 1) are factors of x4 + ax3 − 3x2 + 2x + b.
y3 − 2y2 − 29y − 42
If x + 2 and x − 1 are the factors of x3 + 10x2 + mx + n, then the values of m and n are respectively
If (3x − 1)7 = a7x7 + a6x6 + a5x5 +...+ a1x + a0, then a7 + a5 + ...+a1 + a0 =
If both x − 2 and \[x - \frac{1}{2}\] are factors of px2 + 5x + r, then
Factorise the following:
a4 – 3a2 + 2
