Question

Factorise:
(i) x³ − 3x² + x − 3
(ii) 63x²y² − 7
(iii) 1 − 6x + 9x²
(iv) 7x² − 19x − 6

Answer 1

$$\begin{gathered} \left(\mathrm{i}\right)x^3-3x^2+x-3 \\ =x^2\left(x-3\right)+1\left(x-3\right) \\ =\left(x^2+1\right)\left(x-3\right) \\ \left(\mathrm{ii}\right)63x^2{y}^2-7 \\ =7\left(9x^2{y}^2-1\right) \\ =7\left((3xy{)}^2-(1{)}^2\right)\left[\mathrm{according}\mathrm{to}\mathrm{the}\mathrm{formula}{a}^2-b^2=\left(a+b\right)\left(a-b\right)\right] \\ =7\left(3xy+1\right)\left(3xy-1\right) \\ \left(\mathrm{iii}\right)1-6x+9x^2 \\ =9x^2-6x+1 \\ =9x^2-3x-3x+1 \\ =3x\left(3x-1\right)-1\left(3x-1\right) \\ =\left(3x-1\right)\left(3x-1\right) \\ ={\left(3x-1\right)}^2 \\ \left(\mathrm{iv}\right)7x^2-\; 19x\; -\; 6 \\ =7x^2-21x+2x-6 \\ =7x\left(x-3\right)+2\left(x-3\right) \\ =\left(7x+2\right)\left(x-3\right) \end{gathered}$$