How do you find the roots of x3−12x+16=0 ?

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Fasaniu · Accepted Answer

Explanation:  note that 23−12(2)+16=0  ⇒x=2 is a root and (x−2) is a factor  ⇒x2(x−2)+2x(x−2)−8(x−2)  =(x−2)(x2+2x−8)  =(x−2)(x+4)(x−2)  ⇒(x−2)2(x+4)=0  ⇒x=2 multiplicity of 2  and x=−4 multiplicity of 1.

soanooooo40 · Accepted Answer

Explanation:  Let f(x)=x3−12x+16  Then,  We see that  f(2)=23−12⋅2+16=8−24+16=0  Therefore,  (x−2) is a factor of f(x)  To find the other factors, we perform a long division  x−2| x3+0x2−12x+16| x2+2x−8  x3−2x2  0+2x2−12x  +2x2−4x  +0−8x+16  −8x+16  −8x+16  −0+0  The quotient is  x2+2x−8=(x−2)(x+4)  Therefore,  x3−12x+16=(x−2)2(x+4)

nick1337 · Accepted Answer

Another way.... 12x−x3=16⇒x(12−x2)=16 16 factorises to 16(1)=8(2)=4(4) so we can quickly discover whether or not the function has an integer root. If it does, the factoring is made simple. x=2⇒2(12−4)=2(8) works x=4 gives -16, but x=-4 gives (-4)(-4)=16 x3−12x+16=(x−2)(x+4)(x+c)⇒−8c=16⇒c=−2 Check that this holds. Or (x−2)(x2+bx+c)=x3−12x+16⇒c=−8, bx2−2x2=0⇒b=2 as there is no x2 term.

How do you find the roots of x^3-12x+16=0 ?

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