Question

Solve quadratic equation using factoring

z^2-5z+4=0

Answer 1

`z^2-5z+4=(z-4)(z-1)`

Explanation:

• To factorize a cuadratic equation, one must use the well known formula to find the roots of the equation:
  `(-b(+-)√(b^2-4ac) )/(2a)`, where a (a=1) is the coefficient that accompanies the cuadratic term, b is the coefficient that accompanies the linear term (b=-5) and c is the constant (c=4).

• Then, applied to this specific case, the previous formula results:
  `\frac{5(+-)\sqrt{(-5)^2-4*{1}*{4}} }{2*1}`. Then, the two roots of the equation come from
  `(25(+-)√(9) )/(2)`.
• Consecuently,
  `z_1=(5+√(9))/(2) =4`, and
  `z_2=(5-√(9))/(2) =1`.
• Finally, the cuadratic function can be expressed like this:
  `(z-4)(z-1)` (which comes from the fact that any polynomial can be expressed as the product of
  `(x-x_i)`, where i are the roots of the polynomial). To corroborate that the procedure is correct, you can resolve this product, and you will obtain the inicial expression.