What are the roots of x^3 – 2x^2 – 23x + 60?
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We have to find the roots of : x^3 – 2x^2 – 23x + 60
x^3 – 2x^2 – 23x + 60 = 0
As the equation has a highest power of x, we will have 3 roots.
From the numeric term equal to 60 we know that the product of the roots is 60, substituting values 1,-1, 2, -2 and 3, we get that 3 is a root. So now we have to factor out x – 3, the other roots have a product of 20.
x^3 – 2x^2 – 23x + 60 = 0
(x^3 + x^2 – 20x – 3x^2 – 3x + 60) = 0
(x – 3)(x^2 + x – 20) =0
(x – 3)( x^2 + 5x – 4x – 20) = 0
We can factorize the quadratic term as 5* -4 = -20 and 5 - 4 = 1
(x – 3)( x(x + 5) – 4(x + 5)) = 0
(x – 3)(x – 4)(x + 5) = 0
This gives the roots of x^3 – 2x^2 – 23x + 60 as x = 3 , x = 4 and x = -5.
The required roots are
x = 3 , x = 4 and x = -5.
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To find the roots of x^3 – 2x^2 – 23x + 60.
Let f(x) = x^3-2x^2-23x+60.
We see that f(-5) = (-5)^3-2(-5)^2-23(-5)+60.
f(-5) = -125-50+115+60
f(-5) = -175+175 = 0.
So f(-5) = 0.
Therefore by remainder theorem, if f(0) = 0, then x-a is a factor of f(x).
So f(x) = x^3 – 2x^2 – 23x + 60 has a factor x-(-5) = x+5
Therefore we can write x^3 – 2x^2 – 23x + 60 = (x+5)(x^2+kx+12) as an identity. Put x = 1 in this equation. 1-2-23+60 = (1+5)(1+k+12).
36 = 6(k+13)
36/6 = (k+13)
6 = k+13
So k = 6-13 = -7
Therefore x^3 – 2x^2 – 23x + 60 = (x+5)(x^2-7x+12)
=> x^3 – 2x^2 – 23x + 60 = (x+5)(x-3)(x-4)
So setting each factor to zero, we get the roots :
x+5 = 0, x-3 = 0, x-4 = 0.
So x= -5, x= 3 and x= 4 are the roots.
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