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Step 1:

Factorized the polynomials

5x^{3} - 14x^{2} - 3x = x{5x^{2} - 14x - 3}

= x{5x^{2} - 15x + x - 3}

= x{5x(x - 3) + (x - 3)}

= x(5x + 1)(x - 3)

and

x^{3} - 343 = x^{3} - 7^{3}

= (x - 7)(x^{2} + 7x + 49)

[Using formula, a^{3} - b^{3} = (a - b)(a^{2} + ab + b^{2} )]

Step 2:

=> $\frac{5x^3 - 14x^2 - 3x}{x^3 - 343}$ = $\frac{ x(5x + 1)(x - 3)}{ (x - 7)(x^2 + 7x + 49)}$ is the simplest form of the fraction.

Factorized the polynomials

5x

= x{5x

= x{5x(x - 3) + (x - 3)}

= x(5x + 1)(x - 3)

and

x

= (x - 7)(x

[Using formula, a

Step 2:

=> $\frac{5x^3 - 14x^2 - 3x}{x^3 - 343}$ = $\frac{ x(5x + 1)(x - 3)}{ (x - 7)(x^2 + 7x + 49)}$ is the simplest form of the fraction.

Given quadratic polynomial is 3x^{2} - 7x - 25

Let P(x) = 3x^{2} - 7x - 25

Put x = -2 in 3x^{2} - 7x - 25

=> P(x) = 3(-2)^{2} - (7 * -2) - 25

= 12 + 14 - 25

= 26 - 25

= 1

$\neq$ 0

Hence, x = -2 is not a zero of the quadratic polynomial.

Let P(x) = 3x

Put x = -2 in 3x

=> P(x) = 3(-2)

= 12 + 14 - 25

= 26 - 25

= 1

$\neq$ 0

Hence, x = -2 is not a zero of the quadratic polynomial.

4x

Comparing 4x^{2} + 5x - 9 = 0 with ax^{2} + bx + c = 0

=> a = 4, b = 5 and c = - 9

Now

b^{2} - 4ac = 5^{2} - 4 * 4 * -9

= 25 + 144

= 169

and $\sqrt{b^2 - 4ac}$ = $\sqrt{169}$ = 13

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

= $\frac{-5 \pm13}{2 * 4}$

= $\frac{-5 + 13}{8}$ and $\frac{-5 - 13}{8}$

= $\frac{8}{8}$ and $\frac{-18}{8}$

= 1 and $\frac{-9}{4}$

=> x = 1 and x = $\frac{-9}{4}$.

=> a = 4, b = 5 and c = - 9

Now

b

= 25 + 144

= 169

and $\sqrt{b^2 - 4ac}$ = $\sqrt{169}$ = 13

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

= $\frac{-5 \pm13}{2 * 4}$

= $\frac{-5 + 13}{8}$ and $\frac{-5 - 13}{8}$

= $\frac{8}{8}$ and $\frac{-18}{8}$

= 1 and $\frac{-9}{4}$

=> x = 1 and x = $\frac{-9}{4}$.