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x4 − 4x3 = 6x2 − 12x From least to greatest, what are the integral roots of the equation?

The answer is -2 and 0. I had to take an L. 

Respuesta :

InesWalston

Answer-

The integral roots are -2, 0.

Solution-

The given polynomial is,

[tex]x^4-4x^3-6x^2+12x[/tex]

For calculating the roots,

[tex]\Rightarrow x^4-4x^3-6x^2+12x=0[/tex]

[tex]\Rightarrow x(x^3-4x^2-6x+12)=0[/tex]

Decomposing the terms,

[tex]\Rightarrow x(x^3-6x^2+2x^2+6x-12x+12)=0[/tex]

Rearranging the terms,

[tex]\Rightarrow x(x^3-6x^2+6x+2x^2-12x+12)=0[/tex]

[tex]\Rightarrow x[x(x^2-6x+6)+2(x^2-6x+6)]=0[/tex]

[tex]\Rightarrow x[(x+2)(x^2-6x+6)]=0[/tex]

[tex]\Rightarrow x(x+2)(x^2-6x+6)=0[/tex]

[tex]\Rightarrow x=0,\ x=-2,\ x=\dfrac{-(-6)\pm \sqrt{(-6)^2-4\cdot 1\cdot 6}}{2\cdot 1}[/tex]

[tex]\Rightarrow x=0,\ x=-2,\ x=\dfrac{6\pm \sqrt{12}}{2}[/tex]

[tex]\Rightarrow x=0,\ x=-2,\ x=3\pm \sqrt{3}[/tex]

Therefore, the integral roots are -2, 0.

Answer:

The integral roots of the equation are 0 and -2.

Step-by-step explanation:

Here, the given equation,

[tex]x^4 - 4x^3 = 6x^2 - 12x[/tex]

[tex]x^4-4x^3-6x^2+12x=0[/tex]

[tex]x(x^3-4x^2-6x+12)=0[/tex] ------(1),

Since, [tex]x^3-4x^2-6x+12=0[/tex] at x = - 2,

Thus, x + 2 is a factor of [tex]x^3-4x^2-6x+12=0[/tex],

By dividing [tex]x^3-4x^2-6x+12=0[/tex] by (x+2),

We get, [tex]x^2-6x+6[/tex]

Hence,

[tex]x^3-4x^2-6x+12=(x+2)(x^2-6x+6)[/tex]

By equation (1),

[tex]x(x^3-4x^2-6x+12)=x(x+2)(x^2-6x+6)[/tex]

[tex]\implies x(x+2)(x^2-6x+6)=0[/tex]

[tex]\implies x(x+2)(x-(3+\sqrt{3}))(x-(3-\sqrt{3}))=0[/tex]

If x + 2 = 0 ⇒ x = -2,

While, if x - (3 ± √3 ) = 0 ⇒ x = 3 ± √ 3

Thus the roots of the given equation are, x = 0, - 2, 3 ± √3,

Since, 0 and - 2 are integers,

⇒ The integral roots of the equation are 0 and -2.

   

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