Question

Factorise:

(i) $4a^2-25$ (ii) $x^2-\frac{9}{16}$ (iii) $x^4-y^4$ (iv) $[7\frac{3}{10}{]}^2-[2\frac{1}{10}{]}^2$ (v) $(0.7{)}^2-(0.3{)}^2$ (vi) $(5a-2b{)}^2-(2a-b{)}^2$

Answer 1

(i) $4a^2-25$

$=(2a+5)(2a-5)$ [$\because a^2-b^2=(a-b)(a+b)$]

(ii) $x^2-\frac{9}{16}$

$=(x{)}^2-(\frac{3}{4}{)}^2$

$=(x+\frac{3}{4})(x-\frac{3}{4})$

(iii) $x^4-y^4$

$=(x^2{)}^2-(y^2{)}^2$

$=(x^2+y^2)(x^2-y^2)$

$=(x^2-y^2)(x-y)(x+y)$

(iv) $[7\frac{3}{10}{]}^2-[2\frac{1}{10}{]}^2$

$=[7\frac{3}{10}-2\frac{1}{10}][7\frac{3}{10}+2\frac{1}{10}]$

$=[\frac{73}{10}+\frac{21}{10}][\frac{73}{10}-\frac{21}{10}]$

$=\frac{94}{10}\times \frac{52}{10}$

$=\frac{47}{5}\times \frac{26}{5}$

$=\frac{1222}{25}$

(v) $(0.7{)}^2-(0.3{)}^2$

$=(0.7+0.3)(0.7-0.3)$

$=1\times 0.4$

$=0.4$

(vi) $(5a-2b{)}^2-(2a-b{)}^2$

$=\left[\right(5a-2b)+(2a-b\left)\right]\left[\right(5a-2b)-(2a-b\left)\right]$

$=(7a-3b)(3a-b)$