Question

Question 5
Factorise:
(i) $x^3-2x^2-x+2$
(ii) $x^3-3x^2-9x-5$
(iii) $x^3+13x^2+32x+20$
(iv) $2y^3+y^2-2y-1$

Answer 1

(i) $p\left(x\right)=x^3-2x^2-x+2$
Let us guess a factor (x - a) and choose value of a arbitrarily as 1. Now, putting this value in p(x).
1 - 2 - 1 + 2 = 0
So, (x - 1) is a factor of p(x).
Now, $x^3-2x^2-x+2=x^3-x^2-x^2+x-2x+2=x^2(x-1)-x(x-1)-2(x-1)=(x-1)(x^2-x-2)=(x-1)(x^2-2x+x-2)=(x-1)\left[x(x-2)+1(x-2)\right]=(x-1)(x+1)(x-2)$

(ii) $p\left(x\right)=x^3-3x^2-9x-5$
Take a factor (x - a). a should be a factor of 5, i.e., $\pm$ 1 or $\pm$ 5.
For (x-1), a = 1
$p\left(1\right)=(1{)}^3-(-3)1^2-9\times 1-5=1-3-9-5=-16.$
So, (x - 1) is not a factor of p(x).
For a = 5
$p\left(5\right)=(5{)}^3-3(5{)}^2-9\left(5\right)-5=125-75-45-5=0.$
Therefore, (x - 5) is a factor of $x^3-3x^2-9x-5$
Now, $x^3+3x^2-9x-5=x^3-5x^2+2x^2-10x+x-5=x^2(x-5)+2x(x-5)+1(x-5)=(x-5)(x^2+2x+1)=(x-5)(x+1{)}^2=(x-5\left)\right(x+1\left)\right(x+1)So,x^3-3x^2-9x-5=(x-5\left)\right(x+1\left)\right(x+1)$

(iii) $p\left(x\right)=x^3+13x^2+32x+20$
Let a factor be (x - a). a should be a factor of 20 which arre
$\pm$ 1, $\pm 2,\pm 4,\pm 5,\pm 10.$
For $x-1=0\Rightarrow x=1$
Now, p(1) = 1 + 13 + 32 + 20
= 66 $\ne$ 0
Hence, (x - 1) is not a factor of p(x).
Again, for x + 1 = 0 $\Rightarrow$ x = - 1
Now, p(-1) = - 1 + 13 - 32 + 20
= - 33 + 33 = 0
Hence, (x + 1) is a factors of p(x).
Now, $x^3+13x^2+32x+20=x^3+x^2+12x^2+12x+20x+20=x^2(x+1)+12x(x+1)+20(x+1)=(x+1)(x^2+12x+20)=(x+1)(x^2+10x+2x+20)=(x+1)\left[x\right(x+10)+2(x+10)]=(x+2\left)\right(x+1\left)\right(x+10)$

(iv) $=p\left(y\right)=2y^3+y^2-2y-1$
factors of -2 are $\pm$ 1, $\pm$ 2.
$p\left(1\right)=2\times 1^3+1^2-2\times 1-1=2+1-2-1=0.$
Therefore, (y - 1) is a factor of p(y).
Now, $2y^3+y^2-2y-1=2y^3-2y^2+3y^2-3y+y-1=2y^2(y-1)+3y(y-1)+1(y-1)=(y-1)(2y^2+3y+1)=(y-1)(2y^2+2y+y+1)=(y-1)\left[2y\right(y+1)+1(y+1\left)\right]=(y-1)\left[\right(y+1\left)\right(2y+1\left)\right]$
Therefore, $2y^3+y^2-2y-1=(y-1)(2y+1)(y+1).$