Question

Factorise :
$\left(i\right)16a^2–\frac{25}{4a^2}$
$\left(ii\right)16a^{2}b–\frac{b}{16a^2}$
$\left(iii\right)100(x+y{)}^2–81(a+b{)}^2$
$\left(iv\right)(x–1{)}^2–(x–2{)}^2$

Answer 1

We have,
$\left(i\right)16a^2–\frac{25}{4a^2}$
$=(4a{)}^2-{\left(\frac{5}{2a}\right)}^2=(4a+\frac{5}{2a})(4a-\frac{5}{2a})$
$\left(ii\right)16a^{2}b–\frac{b}{16a^2}=b(16a^2-\frac{1}{16a^2})$
$=b\left\{\right(4a{)}^2-{\left(\frac{1}{4a}\right)}^2\}$
$=b(4a+\frac{1}{4a})(4a-\frac{1}{4a})$

$\left(iii\right)100(x+y{)}^2–81(a+b{)}^2=\left\{10\right(x+y){\}}^2–\{9(a+b){\}}^2$
$=\left\{10\right(x+y)+9(a+b\left)\right\}\left\{10\right(x+y)–9(a+b\left)\right\}$
$=(10x+10y+9a+9b)(10x+10y–9a–9b)$

$\left(iv\right)(x–1{)}^2–(x–2{)}^2=\left\{\right(x–1)+(x–2\left)\right\}\left\{\right(x–1)–(x–2\left)\right\}$
$=(2x–3)(x–1–x+2)$
$=(2x–3)\times 1$
$=2x–3$