Question

Solve this:

​Q1. Factorise

(1) 9x² – 1

(2) 4x⁴ + 12 x³ + 9x²

(3) $\frac{x^2}{9}-\frac{y^2}{25}$
(4) $9x^2+2xy+\frac{y^2}{9}$

Answer 1

$$\begin{gathered} \left(1\right). \\ 9{\mathrm{x}}^2-1 \\ ={\left(3\mathrm{x}\right)}^2-{\left(1\right)}^2 \\ =\left(3\mathrm{x}+1\right)\left(3\mathrm{x}-1\right)\mathbf{}\left[\mathbf{as}\mathbf{,}\mathbf{}{\mathbf{a}}^{\mathbf{2}}\mathbf{-}{\mathbf{b}}^{\mathbf{2}}\mathbf{}\mathbf{=}\mathbf{}\left(\mathbf{a}\mathbf{+}\mathbf{b}\right)\left(\mathbf{a}\mathbf{-}\mathbf{b}\right)\right] \\ \left(\mathbf{2}\right)\mathbf{.} \\ \mathbf{}4{\mathrm{x}}^4+12{\mathrm{x}}^3+9{\mathrm{x}}^2 \\ ={\mathrm{x}}^2\left(4{\mathrm{x}}^2+12\mathrm{x}+9\right) \\ ={\mathrm{x}}^2\left[{\left(2\mathrm{x}\right)}^2+2\times 2\mathrm{x}\times 3+{\left(3\right)}^2\right] \\ ={\mathrm{x}}^2{\left(2\mathrm{x}+3\right)}^2 \\ ={\mathrm{x}}^2\left(2\mathrm{x}+3\right)\left(2\mathrm{x}+3\right)\left[\mathbf{as}\mathbf{,}\mathbf{}{\mathbf{a}}^{\mathbf{2}}\mathbf{}\mathbf{+}\mathbf{}{\mathbf{b}}^{\mathbf{2}}\mathbf{}\mathbf{+}\mathbf{}\mathbf{2}\mathbf{ab}\mathbf{}\mathbf{=}\mathbf{}{\left(\mathbf{a}\mathbf{+}\mathbf{b}\right)}^{\mathbf{2}}\right] \\ \left(3\right). \\ \frac{{\mathrm{x}}^2}{9}-\frac{{\mathrm{y}}^2}{25} \\ ={\left(\frac{\mathrm{x}}{3}\right)}^2-{\left(\frac{\mathrm{y}}{5}\right)}^2 \\ =\left(\frac{\mathrm{x}}{3}+\frac{\mathrm{y}}{5}\right)\left(\frac{\mathrm{x}}{3}-\frac{\mathrm{y}}{5}\right) \\ \left(4\right). \\ 9{\mathrm{x}}^2+2\mathrm{xy}+\frac{{\mathrm{y}}^2}{9} \\ ={\left(3\mathrm{x}\right)}^2+2\left(3\mathrm{x}\right)\left(\frac{\mathrm{y}}{3}\right)+{\left(\frac{\mathrm{y}}{3}\right)}^2 \\ ={\left(3\mathrm{x}+\frac{\mathrm{y}}{3}\right)}^2 \\ =\left(3\mathrm{x}+\frac{\mathrm{y}}{3}\right)\left(3\mathrm{x}+\frac{\mathrm{y}}{3}\right) \end{gathered}$$