Question
Factorise:
x3 - 6x2 + 11x - 6
x3 - 6x2 + 11x - 6
✓
Answer
Let p(x) = x3 - 6x2 + 11x - 6
Constant term of p(x) = -6
$\therefore$ Factors of -6 are ±1, ±2, ±3, ±5, ±6
By trial, we find that p(1) = 0, so (x - 1) is a factor of p(x).
$[\therefore$ (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) + (6x - 1)
= (x - 1)(x2 - 5x + 6) [taking (x - 1) common factor]
Now, (x2 - 5x + 6) = x2 - 3x - 2x + 6 [by spliting middle term]
= x(x - 3) -2(x - 3)
= (x - 3)(x - 2)
$\therefore$ x3 - 6x2 + 11x - 6 = (x - 1)(x - 2)(x - 3)
Constant term of p(x) = -6
$\therefore$ Factors of -6 are ±1, ±2, ±3, ±5, ±6
By trial, we find that p(1) = 0, so (x - 1) is a factor of p(x).
$[\therefore$ (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) + (6x - 1)
= (x - 1)(x2 - 5x + 6) [taking (x - 1) common factor]
Now, (x2 - 5x + 6) = x2 - 3x - 2x + 6 [by spliting middle term]
= x(x - 3) -2(x - 3)
= (x - 3)(x - 2)
$\therefore$ x3 - 6x2 + 11x - 6 = (x - 1)(x - 2)(x - 3)
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Explore more
Similar questions
Simplify:
$\Big(\frac{\sqrt{2}}{5}\Big)^8\div\Big(\frac{\sqrt{2}}{5}\Big)^{13}$
→$\Big(\frac{\sqrt{2}}{5}\Big)^8\div\Big(\frac{\sqrt{2}}{5}\Big)^{13}$
Express the following decimals in the form $\frac{\text{p}}{\text{q}},$ where p, q are integers and $\text{q}\neq0.$
$0.\overline{235}$
→$0.\overline{235}$
In figure, $\angle{1}=60^\circ$ and $\angle{2}=\Big(\frac{2}{3}\Big)^\text{rd}$ of a right angle. Prove that l∥m.
→Find the irrational number between $\frac{1}{7}$and $\frac{2}{7}.$
→The exterior angles, obtained on producing the base of a triangle both ways are 104° and 136°. Find all the angles of the triangle.
Prove that:
$\frac{\text{abc}}{\text{a}^{-1}\text{b}^{-1}+\text{b}^{-1}\text{c}^{-1}+\text{c}^{-1}\text{a}^{-1}}=\text{abc}$
→$\frac{\text{abc}}{\text{a}^{-1}\text{b}^{-1}+\text{b}^{-1}\text{c}^{-1}+\text{c}^{-1}\text{a}^{-1}}=\text{abc}$
Rationalise the denominator of the following:
$\frac{2+\sqrt{3}}{2-\sqrt{3}}$
→$\frac{2+\sqrt{3}}{2-\sqrt{3}}$
Find five rational numbers between $\frac{3}{5}$ and $\frac{4}{5}.$
→An angle is equal to its supplement. Determine its measure.
The height of a cone 21cm. Find the area of the base if the slant height is 28cm.