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### Solved Examples

**Question 1:**Simplify $\frac{5x^3 - 14x^2 - 3x}{x^3 - 343}$

**Solution:**

Step 1:

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.

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.

**Question 2:**Show that x = -2 is a zero of quadratic polynomial 3x

^{2}- 7x - 25.

**Solution:**

Given quadratic polynomial is 3x

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.

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

^{2}- 7x - 25Put 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.

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

4x

^{2}+ 5x - 9 = 0

**Solution:**

Comparing 4x

=> 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 = $\frac{-9}{4}$.

^{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 = $\frac{-9}{4}$.