What is the factored expression of 63x^2-3x-6

Answer:

$3(3x-1)(7x +2)$

Step-by-step explanation:

$63x^2-3x-6$

Suppose a generic quadratic equation

$ax^2 + bx + c$

To factor this equation, I need to find its roots. Then we use the quadratic formula of:

$\frac{-b + \sqrt{b^2-4ac}}{2a}$

and

$\frac{-b + \sqrt{b^2-4ac}}{2a}$

So, for the equation 63x ^ 2-3x-6 we have:

$\frac{3 + \sqrt{(-3)^2-4(63)(-6)}}{2(63)} = \frac{1}{3}$

and

$\frac{3 - \sqrt{(-3)^2-4(63)(-6)}}{2(63)} = \frac{-2}{7}$

So:

$(x-\frac{1}{3}) = 0\\\\(3x-1) = 0$

and

$(x - (-\frac{2}{7})) = 0\\\\(x+ \frac{2}{7}) = 0\\\\(7x +2) = 0$

Finally the polynomial is:

$(3x-1)(7x +2)$

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