Answer:
[tex]z^2-5z+4=(z-4)(z-1)[/tex]
Step-by-step explanation:
- To factorize a cuadratic equation, one must use the well known formula to find the roots of the equation: [tex]\frac{-b(+-)\sqrt{b^2-4ac} }{2a}[/tex], 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: [tex]\frac{5(+-)\sqrt{(-5)^2-4\times{1}\times{4}} }{2\times1}[/tex]. Then, the two roots of the equation come from [tex]\frac{25(+-)\sqrt{9} }{2}[/tex].
- Consecuently, [tex]z_1=\frac{5+\sqrt{9}}{2} =4[/tex], and [tex]z_2=\frac{5-\sqrt{9}}{2} =1[/tex].
- Finally, the cuadratic function can be expressed like this: [tex](z-4)(z-1)[/tex] (which comes from the fact that any polynomial can be expressed as the product of [tex](x-x_i)[/tex], 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.
