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Given $\frac{45x^{12}}{30x^6}$

Step 1:

Factorized the numerator and denominator

45 x^{12} = 3 * 3 * 5 * x^{12}

30 x^{6} = 2 * 3 * 5 * x^{6}

Step 2:

$\frac{45x^{12}}{30x^6}$ = $\frac{3 * 3 * 5 * x^{12}}{ 2 * 3 * 5 * x^6}$

= $\frac{3 * x^{12}}{ 2 * x^6}$

= $\frac{3}{2}$$ x^{12 - 6}$

[ $\frac{x^m}{x^n}$ = x^{m - n} ]

= $\frac{3}{2}$$ x^6$

=> $\frac{45x^{12}}{30x^6}$ = $\frac{3}{2}$$ x^6$

Step 1:

Factorized the numerator and denominator

45 x

30 x

Step 2:

$\frac{45x^{12}}{30x^6}$ = $\frac{3 * 3 * 5 * x^{12}}{ 2 * 3 * 5 * x^6}$

= $\frac{3 * x^{12}}{ 2 * x^6}$

= $\frac{3}{2}$$ x^{12 - 6}$

[ $\frac{x^m}{x^n}$ = x

= $\frac{3}{2}$$ x^6$

=> $\frac{45x^{12}}{30x^6}$ = $\frac{3}{2}$$ x^6$

2x

Given quadratic equation 2x^{2} + 5x - 3 = 0

Comparing with equation ax^{2} + bx + c = 0

a = 2, b = 5 and c = - 3

b^{2} - 4ac = 5^{2} - 4 * 2 * - 3

= 25 + 24

= 49

and $\sqrt{b^2 - 4ac}$ = $\sqrt{49}$

= 7

=> $\sqrt{b^2 - 4ac}$ = 7

therefore

x = $\frac{- b \pm \sqrt{b^2 - 4ac}}{2a}$

= $\frac{- 5 \pm 7}{2 * 2}$

= $\frac{- 5 \pm 7}{4}$

=> x = $\frac{- 5 + 7}{4}$ or x = $\frac{- 5 - 7}{4}$

=> x = $\frac{1}{2}$, - 3.

Comparing with equation ax

a = 2, b = 5 and c = - 3

b

= 25 + 24

= 49

and $\sqrt{b^2 - 4ac}$ = $\sqrt{49}$

= 7

=> $\sqrt{b^2 - 4ac}$ = 7

therefore

x = $\frac{- b \pm \sqrt{b^2 - 4ac}}{2a}$

= $\frac{- 5 \pm 7}{2 * 2}$

= $\frac{- 5 \pm 7}{4}$

=> x = $\frac{- 5 + 7}{4}$ or x = $\frac{- 5 - 7}{4}$

=> x = $\frac{1}{2}$, - 3.

Given, $\frac{x^3 - 2^3}{x^2 + 2x - 8}$

Step 1:

x^{3} - 2^{3} = (x - 2)(x^{2} + 2x + 2^{2} )

= (x - 2)(x^{2} + 2x + 4)

[x^{3} - y^{3} = (x - y)(x^{2} + xy + y^{2} )]

and

x^{2} + 2x - 8 = x^{2} + 4x - 2x - 8

= x(x + 4) - 2(x + 4)

= (x + 4)(x - 2)

Step 2:

$\frac{x^3 - 2^3}{x^2 + 2x - 8}$ = $\frac{ (x - 2)(x^2 + 2x + 4)}{ (x + 4)(x - 2) }$

= $\frac{x^2 + 2x + 4}{x + 4}$

=> $\frac{x^3 - 2^3}{x^2 + 2x - 8}$ = $\frac{x^2 + 2x + 4}{x + 4}$

Step 1:

x

= (x - 2)(x

[x

and

x

= x(x + 4) - 2(x + 4)

= (x + 4)(x - 2)

Step 2:

$\frac{x^3 - 2^3}{x^2 + 2x - 8}$ = $\frac{ (x - 2)(x^2 + 2x + 4)}{ (x + 4)(x - 2) }$

= $\frac{x^2 + 2x + 4}{x + 4}$

=> $\frac{x^3 - 2^3}{x^2 + 2x - 8}$ = $\frac{x^2 + 2x + 4}{x + 4}$