Answer:
[tex]3(3x-1)(7x +2)[/tex]
Step-by-step explanation:
[tex]63x^2-3x-6[/tex]
Suppose a generic quadratic equation
[tex]ax^2 + bx + c[/tex]
To factor this equation, I need to find its roots. Then we use the quadratic formula of:
[tex]\frac{-b + \sqrt{b^2-4ac}}{2a}[/tex]
and
[tex]\frac{-b + \sqrt{b^2-4ac}}{2a}[/tex]
So, for the equation 63x ^ 2-3x-6 we have:
[tex]\frac{3 + \sqrt{(-3)^2-4(63)(-6)}}{2(63)} = \frac{1}{3}[/tex]
and
[tex]\frac{3 - \sqrt{(-3)^2-4(63)(-6)}}{2(63)} = \frac{-2}{7}[/tex]
So:
[tex](x-\frac{1}{3}) = 0\\\\(3x-1) = 0[/tex]
and
[tex](x - (-\frac{2}{7})) = 0\\\\(x+ \frac{2}{7}) = 0\\\\(7x +2) = 0[/tex]
Finally the polynomial is:
[tex](3x-1)(7x +2)[/tex]