Question

Q. Factorise the following :
$$\begin{gathered} i.2x^2-7x-15ii.84-2r-2r^2 \\ iii.2x^2-\frac{5}{6}x+\frac{1}{12}iv.3a^{3}b-243a{b}^3 \\ v.x^4-625vi.a^7-a{b}^6 \\ vii.x^{12}-y^{12} \end{gathered}$$

Answer 1

Dear student
$$\begin{gathered} \mathrm{i} \\ 2{\mathrm{x}}^2-7\mathrm{x}-15 \\ =2{\mathrm{x}}^2-10\mathrm{x}+3\mathrm{x}-15 \\ =2\mathrm{x}\left(\mathrm{x}-5\right)+3\left(\mathrm{x}-5\right) \\ =\left(2\mathrm{x}+3\right)\left(\mathrm{x}-5\right) \\ \left(\mathrm{viii}\right) \\ {\mathrm{x}}^{12}-{\mathrm{y}}^{12} \\ ={\left({\mathrm{x}}^6\right)}^2-{\left({\mathrm{y}}^6\right)}^2 \\ =\left({\mathrm{x}}^6+{\mathrm{y}}^6\right)\left({\mathrm{x}}^6-{\mathrm{y}}^6\right) \\ =\left({\mathrm{x}}^6+{\mathrm{y}}^6\right){\left({\mathrm{x}}^3\right)}^2-{\left({\mathrm{y}}^3\right)}^2 \\ =\left({\mathrm{x}}^6+{\mathrm{y}}^6\right)\left({\mathrm{x}}^3-{\mathrm{y}}^3\right)\left({\mathrm{x}}^3+{\mathrm{y}}^3\right) \\ =\left({\mathrm{x}}^6+{\mathrm{y}}^6\right)\left(\mathrm{x}-\mathrm{y}\right)\left({\mathrm{x}}^2+{\mathrm{y}}^2+\mathrm{xy}\right)\left(\mathrm{x}+\mathrm{y}\right)\left({\mathrm{x}}^2+{\mathrm{y}}^2-\mathrm{xy}\right) \\ =\left(\mathrm{x}+\mathrm{y}\right)\left(\mathrm{x}-\mathrm{y}\right)\left({\mathrm{x}}^6+{\mathrm{y}}^6\right)\left({\mathrm{x}}^2+{\mathrm{y}}^2+\mathrm{xy}\right)\left({\mathrm{x}}^2+{\mathrm{y}}^2-\mathrm{xy}\right) \\ \mathrm{Note}: \\ {\mathrm{a}}^2-{\mathrm{b}}^2=\left(\mathrm{a}+\mathrm{b}\right)\left(\mathrm{a}-\mathrm{b}\right) \\ {\mathrm{a}}^3+{\mathrm{b}}^3=\left(\mathrm{a}+\mathrm{b}\right)\left({\mathrm{a}}^2+{\mathrm{b}}^2-\mathrm{ab}\right) \\ {\mathrm{a}}^3-{\mathrm{b}}^3=\left(\mathrm{a}-\mathrm{b}\right)\left({\mathrm{a}}^2+{\mathrm{b}}^2+\mathrm{ab}\right) \end{gathered}$$
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