Question
Factorise:
$x^3-3 x^2+3 x+7$
$x^3-3 x^2+3 x+7$
✓
Answer
$x^3-3 x^2+3 x+7$
$=x^3-3 x^2+3 x-1+8$
$=\left(x^3-3 x^2+3 x-1\right)+8$
$=(x-1)^3+2^3$
$=(x-1+2)\left[(x-1)^2-(x-1)(2)+2^2\right]$
$=(x+1)\left(x^2-2 x+1-2 x+2+4\right)$
$=(x+1)\left(x^2-4 x+7\right)$
$=x^3-3 x^2+3 x-1+8$
$=\left(x^3-3 x^2+3 x-1\right)+8$
$=(x-1)^3+2^3$
$=(x-1+2)\left[(x-1)^2-(x-1)(2)+2^2\right]$
$=(x+1)\left(x^2-2 x+1-2 x+2+4\right)$
$=(x+1)\left(x^2-4 x+7\right)$
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
Factorise:
$a^3-8 b^3-64 c^3-24 a b c$
→$a^3-8 b^3-64 c^3-24 a b c$
If a line intersects two concentric circles $($circles with the same centre$)$ with centre $O$ at $A , B , C$ and $D ,$ prove that $A B=C D$.
→Find the remainder when $ {x^3} + 3{x^2} + 3x + 1$ is divided by $x - \frac{1}{2}$
→Factorize: $\text{x}^2+5\sqrt{5}\text{x}+30$
→Show that $0.2353535... = 0. 2\overline{35}$, can be expressed in the form $\frac{p}{q}$, where $p$ and $q$ are integers and q $\ne 0.$
→Factorise:
$1-27 a^3$
→$1-27 a^3$
By Remainder Theorem find the remainder, when p(x) is divided by $g(x)$, where: $\text{p(x)}=\text{x}^3-6\text{x}^2+2\text{x}-4,$ and $\text{g(x)}=1-\frac{3}{2}\text{x} $
→The percentages of marks obtained by a student in six unit tests are given below:
A unit test is selected at random. What is the probability that the student gets more than 60% marks in the test?
→|
Unit test
|
$I$
|
$II$
|
$III$
|
$IV$
|
$V$
|
$VI$
|
|
Percentage of marks obtained
|
$53$
|
$72$
|
$28$
|
$46$
|
$67$
|
$59$
|
Factorise:
$8 a^3+125 b^3-64 c^3+120 a b c$
→$8 a^3+125 b^3-64 c^3+120 a b c$
Simplify: $\left[5\left(8^{\frac{1}{3}}+27^{\frac{1}{3}}\right)^3\right]^{\frac{1}{4}}$
→