Question

factorise the following using algebraic identifies​

Answer 1

**GIVEn**

• a⁴ + 6a²b² + 9b⁴
• x² + 1/x² + 2
• x² + x + 1/4

Factories it

**SOLUTIOn**

• a⁴ + 6a²b² + 9b⁴

Apply identity :

(a + b)² = a² + b² + 2ab

→ a⁴ + 9b⁴ + 6a²b²

→ (a²)² + (3b²)² + 2 × a² × 3b²

→ (a² + b²)²

→ (a² + b²)(a² + b²)

_____________________

• x² + 1/x² + 2

Apply identity :

Apply identity :(a + b)² = a² + b² + 2ab

→ (x)² + (1/x)² + 2 × x × 1/x

→ (x + 1/x)²

→ (x + 1/x)(x + 1/x)

_____________________

• x² + x + 1/4

Apply identity :

Apply identity :(a + b)² = a² + b² + 2ab

→ x² + 1/4 + x

→ (x)² + (1/2)² + 2 × x × 1/2

→ (x + 1/2)²

→ (x + 1/2)(x + 1/2)

_____________________

Answer 2

Answer:

$</p><p>1. a^4 +6a^2b^2 +9b^4 \\ \\ \\</p><p>Identity - (a+b)^2 = a^2+b^2 +2ab\\ \\ \\</p><p>(a^2)^2 +2(a^2× 3b^2) +(3b^2)^2 \\ \\ \\</p><p>= (a^2 + 3b ^2)^2 \\ \\ \\</p><p>2. x^2 +\frac {1}{x^2}+2\\ \\ \\</p><p>using \: the \: same \: identity \: as \: above \\ \\ \\</p><p>(x)^2 +(\frac{1}{x})^2 +2(x × \frac{1}{x}) \\ \\ \\</p><p>x \:and\: \frac{1}{x} \: gets\: cancelled \: so \: the \\ \\ \\last \: term \: becomes \: 2\\ \\ \\</p><p>= (x+\frac{1}{x})^2 \\ \\ \\</p><p>3. x^2+x+\frac{1}{4}\\ \\ \\</p><p>using\: same\: identity\\ \\ \\</p><p>(x)^2 +2(x× \frac{1}{2})+(\frac{1}{2})^2 \\ \\ \\</p><p>2× \frac{x}{2}\: becomes \: x \: which \: is \: the \\ \\ \\middle \: term\\ \\ \\</p><p>= (x+\frac{1}{2})^2$