# Equations- Factorization of a polynomial equation – Exercise 5652

Exercise

Factor the polynomial equation

$$x^3-4x^2+5=0$$

$$(x-\frac{5+ \sqrt{5}}{2})(x-\frac{5- \sqrt{5}}{2})(x+1)=0$$

Solution

$$x^3-4x^2+5=$$

$$=x^3-5x^2+x^2+5x-5x+5=$$

$$=x^3-5x^2+5x+x^2-5x+5=$$

$$=x(x^2-5x+5)+(x^2-5x+5)=$$

$$=(x^2-5x+5)(x+1)=$$

The first factor is a quadratic equation. its coefficients are:

$$a=1, b=-5, c=5$$

We solve it with the quadratic formula. Putting the coefficients in the formula gives us

$$x_{1,2}=\frac{5\pm \sqrt{{(-5)}^2-4\cdot 1\cdot 5}}{2\cdot 1}=$$

$$=\frac{5\pm \sqrt{5}}{2}$$

Hence, we get the solutions:

$$x_1=\frac{5+ \sqrt{5}}{2}$$

$$x_2=\frac{5- \sqrt{5}}{2}$$

Thus, the factorizing of the quadratic equation is

$$x^2-5x+5=$$

$$=(x-\frac{5+ \sqrt{5}}{2})(x-\frac{5- \sqrt{5}}{2})$$

All together we get

$$x^3-4x^2+5=$$

$$(x-\frac{5+ \sqrt{5}}{2})(x-\frac{5- \sqrt{5}}{2})(x+1)$$

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