Exercise
Factor the polynomial equation
x 3 − 4 x 2 + 5 = 0 x^3-4x^2+5=0 x 3 − 4 x 2 + 5 = 0
Final Answer
Show final answer
( x − 5 + 5 2 ) ( x − 5 − 5 2 ) ( x + 1 ) = 0 (x-\frac{5+ \sqrt{5}}{2})(x-\frac{5- \sqrt{5}}{2})(x+1)=0 ( x − 2 5 + 5 ) ( x − 2 5 − 5 ) ( x + 1 ) = 0
Solution
x 3 − 4 x 2 + 5 = x^3-4x^2+5= x 3 − 4 x 2 + 5 =
= x 3 − 5 x 2 + x 2 + 5 x − 5 x + 5 = =x^3-5x^2+x^2+5x-5x+5= = x 3 − 5 x 2 + x 2 + 5 x − 5 x + 5 =
= x 3 − 5 x 2 + 5 x + x 2 − 5 x + 5 = =x^3-5x^2+5x+x^2-5x+5= = x 3 − 5 x 2 + 5 x + x 2 − 5 x + 5 =
= x ( x 2 − 5 x + 5 ) + ( x 2 − 5 x + 5 ) = =x(x^2-5x+5)+(x^2-5x+5)= = x ( x 2 − 5 x + 5 ) + ( x 2 − 5 x + 5 ) =
= ( x 2 − 5 x + 5 ) ( x + 1 ) = =(x^2-5x+5)(x+1)= = ( x 2 − 5 x + 5 ) ( x + 1 ) =
The first factor is a quadratic equation
a = 1 , b = − 5 , c = 5 a=1, b=-5, c=5 a = 1 , b = − 5 , c = 5
We solve it with the quadratic formula
x 1 , 2 = 5 ± ( − 5 ) 2 − 4 ⋅ 1 ⋅ 5 2 ⋅ 1 = x_{1,2}=\frac{5\pm \sqrt{{(-5)}^2-4\cdot 1\cdot 5}}{2\cdot 1}= x 1 , 2 = 2 ⋅ 1 5 ± ( − 5 ) 2 − 4 ⋅ 1 ⋅ 5 =
= 5 ± 5 2 =\frac{5\pm \sqrt{5}}{2} = 2 5 ± 5
Hence, we get the solutions:
x 1 = 5 + 5 2 x_1=\frac{5+ \sqrt{5}}{2} x 1 = 2 5 + 5
x 2 = 5 − 5 2 x_2=\frac{5- \sqrt{5}}{2} x 2 = 2 5 − 5
Thus, the factorizing of the quadratic equation
x 2 − 5 x + 5 = x^2-5x+5= x 2 − 5 x + 5 =
= ( x − 5 + 5 2 ) ( x − 5 − 5 2 ) =(x-\frac{5+ \sqrt{5}}{2})(x-\frac{5- \sqrt{5}}{2}) = ( x − 2 5 + 5 ) ( x − 2 5 − 5 )
All together we get
x 3 − 4 x 2 + 5 = x^3-4x^2+5= x 3 − 4 x 2 + 5 =
( x − 5 + 5 2 ) ( x − 5 − 5 2 ) ( x + 1 ) (x-\frac{5+ \sqrt{5}}{2})(x-\frac{5- \sqrt{5}}{2})(x+1) ( x − 2 5 + 5 ) ( x − 2 5 − 5 ) ( x + 1 )
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