Find the solutions to x^3+4x^2+x-5=1

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Find the solutions to x^3+4x^2+x-5=1

These cubic equations are usually fairly simple once the method is known, firstly make the right hand side (RHS) equal to zero by subtracting 1 from both sides:

x^3+4x^2+x-6=0

Now the next step is a quick bit of trial and error, one of the roots to this equation is a factor of 6 so this leaves us with ±1, ±2, ±3, ±6. So plug these into the equation until you get a result which equals zero.

After trialling we find that +1 is a root or in factorised form (x-1). This leads us to our next step:

What quadratic multiplied by our factorised root gives us our original equation?

(x-1)(ax^2+bx+c)=x^3+4x^2+x-6

Well the coefficient of x^3 is 1 so our value of 'a' must also be 1 (If the coeffiecient of x^3 was 5 then 'a' would also be 5)

(x-1)(x^2+bx+c)=x^3+4x^2+x-6

The constant term 'c' must equal -6 when multiplied by -1 and so c= (-6/-1) =6

(x-1)(x^2+bx+6)=x^3+4x^2+x-6

Now we need to find b...

The coefficient of x in the problem was 1, so we need to find a value of 'b' so that when the brackets are expanded there is 1x:

1x=bx*(-1)+x*6 so b=5

This gives us:

(x-1)(x^2+5x+6)=x^3+4x^2+x-6

(x-1)(x^2+5x+6)=0

Factorise the quadratic either by inspection or by using the quadratic formula:

(x-1)(x+2)(x+3)=0

so the roots are x=1,-2,-3

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