Question
Factorise:
$x^3+x^2-4 x-4$
$x^3+x^2-4 x-4$
✓
Answer
Let $p(x)=x^3+x^2-4 x-4$
Constant term of $p(x)=-4$
$\therefore$ Factors of $-4$ are $\pm 1, \pm 2, \pm 4$
By trial, we find that $p(-1)=0$, so $(x+1)$ is a factor of $p(x)$.
Now, we see that $x^3+x^2-4 x-4$
$=x^2(x+1)-4(x+1)$
$=(x+1)\left(x^2-4\right)[\text { taking }(x+1) \text { common factor] }$
Now, $x^2-4=x^2-2^2$
$=(x+2)(x-2)\left[\text { Using identity, } a^2-b^2=(a-b)(a+b)\right]$
$\therefore x^3+x^2-4 x-4=(x+1)(x-2)(x+2)$
Constant term of $p(x)=-4$
$\therefore$ Factors of $-4$ are $\pm 1, \pm 2, \pm 4$
By trial, we find that $p(-1)=0$, so $(x+1)$ is a factor of $p(x)$.
Now, we see that $x^3+x^2-4 x-4$
$=x^2(x+1)-4(x+1)$
$=(x+1)\left(x^2-4\right)[\text { taking }(x+1) \text { common factor] }$
Now, $x^2-4=x^2-2^2$
$=(x+2)(x-2)\left[\text { Using identity, } a^2-b^2=(a-b)(a+b)\right]$
$\therefore x^3+x^2-4 x-4=(x+1)(x-2)(x+2)$
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
Draw the graph of x + y = 7
In fig lines $XY$ and $MN$ intersect at $O$ If $\angle POY = 90$^\circ and $a:b = 2:3$ find $\angle c.$
→A spherical cannonball $28\ cm$ in diameter is melted and cast into a right circular cone mould, whose base is $35\ cm$ in diameter. Find the height of the cone.
→A circus tent is cylindrical to a height of $3m$ and conical above it. If its diameter is $105m$ and the slant height of the conical portion is $53m$, calculate the length of the canvas $5m$ wide to make the required tent.
→$ABCD$ is a parallelogram in which $\angle\text{A}=70^\circ$ Compute $\angle\text{B},\angle\text{C}$ and $\angle\text{D}$.
→Express the following decimals in the form $\frac{\text{p}}{\text{q}}:$ $0.\overline{621}$
→Given below are the seats won by different political parties in the polling oucome of a state assembly elections:
Draw a bar graph to represent the polling results.
→|
Political party
|
$A$
|
$B$
|
$C$
|
$D$
|
$E$
|
$F$
|
|
Seats won
|
$65$
|
$52$
|
$34$
|
$28$
|
$10$
|
$31$
|
Prove that two different circles cannot intersect each other at more than two points.
It being given that $\sqrt{3}=1.732,\sqrt{5}=2.236,\sqrt{6}=2.449$ and $\sqrt{10}=3.162,$ find to three places of decimal, the value of the following: $\frac{1+2\sqrt{3}}{2-\sqrt{3}}$
→Simplify by rationalising the denominator: $\frac{2\sqrt{6}-\sqrt{5}}{3\sqrt{5}-2\sqrt{6}}$
→