# Help with Math Problems

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## Solved Examples

Question 1: Simplify $\frac{5x^3 - 14x^2 - 3x}{x^3 - 343}$
Solution:
Step 1:
Factorized the polynomials

5x3 - 14x2 - 3x = x{5x2 - 14x - 3}

= x{5x2 - 15x + x - 3}

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

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

and

x3 - 343 = x3 - 73

= (x - 7)(x2 + 7x + 49)

[Using formula,  a3 - b3 = (a - b)(a2 + ab + b2 )]

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.

Question 2: Show that x = -2 is a zero of quadratic polynomial 3x2 - 7x - 25.

Solution:
Given quadratic polynomial is 3x2 - 7x - 25

Let P(x) = 3x2 - 7x - 25

Put x = -2 in 3x2 - 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.

Question 3: Solve the quadratic equation by using the formula.

4x2 + 5x - 9 = 0
Solution:
Comparing 4x2 + 5x - 9 = 0 with ax2 + bx + c = 0

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

Now
b2 - 4ac = 52 - 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 = $\frac{-9}{4}$.