Question
Factorise:$10\Big(3\text{x}+\frac{1}{\text{x}}\Big)^2-\Big(3\text{x}+\frac{1}{\text{x}}\Big)-3$
✓
Answer
Given equation:
$10\Big(3\text{x}+\frac{1}{\text{x}}\Big)^2-\Big(3\text{x}+\frac{1}{\text{x}}\Big)-3$ Let $3\text{x}+\frac{1}{\text{x}}=\text{a}$
Then, we have $= 10a^2 - a - 3 = 10a^2 - 6a + 5a - 3$
$= 2a(5a - 3) + 1(5a - 3) = (5a - 3)(2a + 1)$
$=\bigg[5\Big(3\text{x}+\frac{1}{\text{x}}\Big)-3\bigg]\bigg[2\Big(3\text{x}+\frac{1}{\text{x}}\Big)+1\bigg]$
$=\Big(15\text{x}+\frac{5}{\text{x}}-3\Big)\Big(6\text{x}+\frac{2}{\text{x}}+1\Big)$
$10\Big(3\text{x}+\frac{1}{\text{x}}\Big)^2-\Big(3\text{x}+\frac{1}{\text{x}}\Big)-3$ Let $3\text{x}+\frac{1}{\text{x}}=\text{a}$
Then, we have $= 10a^2 - a - 3 = 10a^2 - 6a + 5a - 3$
$= 2a(5a - 3) + 1(5a - 3) = (5a - 3)(2a + 1)$
$=\bigg[5\Big(3\text{x}+\frac{1}{\text{x}}\Big)-3\bigg]\bigg[2\Big(3\text{x}+\frac{1}{\text{x}}\Big)+1\bigg]$
$=\Big(15\text{x}+\frac{5}{\text{x}}-3\Big)\Big(6\text{x}+\frac{2}{\text{x}}+1\Big)$
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