CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon


Question

Given that $$x-  \sqrt{5}$$ is a factor of the cubic polynomial $$x^3 -3 \sqrt{5}x^2 + 13x -3 \sqrt{5}$$ , find all the zeroes of the polynomial.


A
5,5+2,52
loader
B
5,5,52
loader
C
5,5+2,5
loader
D
None of the above
loader

Solution

The correct option is A $$\sqrt{5}, \sqrt{5}+ \sqrt{2}, \sqrt{5} -\sqrt{2}$$
If $$(x-\sqrt { 5 })$$ is a factor, then we can write: 

$$x^3–3\sqrt { 5 } x^2+13x–3\sqrt { 5 } =(x–\sqrt { 5 } )(x^2+bx+3)$$ 
  
To determine the coefficient $$b$$, let's expand the product: 

$$(x–\sqrt { 5 } )(x^2+bx+3)=x^3+bx^2+3x–(\sqrt { 5 } )x^2–(\sqrt { 5 } )bx–3\sqrt { 5 }$$

$$(x–\sqrt { 5 } )(x^2+bx+3)=x^3+(b–\sqrt { 5 } )x^2+(3–b\sqrt { 5 } )x–3\sqrt { 5 }$$  

Comparing the right hand side to the original expression, we obtain 

$$b–\sqrt { 5 } =-3\sqrt { 5 } ⇒b=-2\sqrt { 5 }$$, or, with the same result: 

$$3–b\sqrt { 5 } =13\\ ⇒b\sqrt { 5 } =-10\\ ⇒b=-10/\sqrt { 5 } =-2\sqrt { 5 } \\ ⇒b=-2\sqrt { 5 }$$  

Therefore,
 
$$x^3–3\sqrt { 5 } x^2+13x–3\sqrt { 5 } =(x–\sqrt { 5 } )(x^2–2\sqrt { 5 } x+3)\\ x^3–3\sqrt { 5 } x^2+13x–3\sqrt { 5 } =0\\ (x–\sqrt { 5 } )=0,\quad (x^2–2\sqrt { 5 } x+3)=0\\ x–\sqrt { 5 } =0⇒x=\sqrt { 5 } \\ x^2–2\sqrt { 5 } x+3=0⇒x=\sqrt { 5 } ±\sqrt { 2 }$$ 

Hence, the zeros of the given expression are $$\sqrt { 5 } +\sqrt { 2 } ,\sqrt { 5 } -\sqrt { 2 } ,\sqrt { 5 }$$.

Mathematics

Suggest Corrections
thumbs-up
 
0


similar_icon
Similar questions
View More


similar_icon
People also searched for
View More



footer-image