Question

`x^3+4x^2+4x+3` Please solve it in a simplest form with proper method????

Answer 1

`x^3+4x^2+4x+3 = (x+3)(x^2+x+1)`

`color(white)(x^3+4x^2+4x+3) = (x+3)(x+1/2-sqrt(3)/2i)(x+1/2+sqrt(3)/2i)`

Explanation:

Note that:

`x^3+4x^2+4x+3 = (x^3+x^2+x)+(3x^2+3x+3)`

`color(white)(x^3+4x^2+4x+3) = x(x^2+x+1)+3(x^2+x+1)`

`color(white)(x^3+4x^2+4x+3) = (x+3)(x^2+x+1)`

The remaining quadratic is recognisable as `(x^3-1)/(x-1)`, but let's just factor it by completing the square:

`x^2+x+1 = (x+1/2)^2+3/4`

`color(white)(x^2+x+1) = (x+1/2)^2+(sqrt(3)/2)^2`

`color(white)(x^2+x+1) = (x+1/2)^2-(sqrt(3)/2i)^2`

`color(white)(x^2+x+1) = ((x+1/2)-sqrt(3)/2i)((x+1/2)+sqrt(3)/2i)`

`color(white)(x^2+x+1) = (x+1/2-sqrt(3)/2i)(x+1/2+sqrt(3)/2i)`

So:

`x^3+4x^2+4x+3 = (x+3)(x^2+x+1)`

`color(white)(x^3+4x^2+4x+3) = (x+3)(x+1/2-sqrt(3)/2i)(x+1/2+sqrt(3)/2i)`

Answer 2

`(x+3)(x+1/2+(isqrt3)/2)(x+1/2-(isqrt3)/2)`

Explanation:

If you want this to be factored in simplest form...
you can try to use the rational zero theorem.

We have:

`x^3+4x^2+4x+3`

Instead of explaining the theorem, I will just explain the steps.

First, identify the constant and the coefficient of the variable that is raised to the highest power.

The constant is `3` and the coefficient is `1`.

Now, factorize `3`.

`3=>1*3 or -1*-3`

Now, factorize `1.`

`1=>1 or -1*-1` (Of course, you can go on like this infinitely on.)

Now, divide the factors of the constants by the factors of the coefficient. Those are the possible zeros.

The possible zeros are:

`+-1,+-3`

Plug each possibility. If a possibility gives you a zero, then it is a zero for the function.

In our case, `-3` does.

Now, we need the synthetic division.

First, write out the coefficients of the polynomial:

The result you get here (`1,1,1,0` in our case) is the coefficients for your new polynomial (The second to last one is the constant).

This new polynomial is a degree lower than our original one.

We now have:

`x^2+x+1` We now let this equal zero and solve for `x`.

`x^2+x+1=0` Use the quadratic formula.

`x=(-b+-sqrt(b^2-4(a)(c)))/(2(a))`

`=>x=(-1+-sqrt(1^2-4(1)(1)))/(2(1))`

`=>x=(-1+-sqrt(1-4))/2`

`=>x=(-1+-sqrt(-3))/2`

`=>x=(-1+-isqrt(3))/2`

`=>(-1-isqrt(3))/2=x=(-1+isqrt(3))/2` These two are the other zeros.

The zeros are:

`-3,(-1-isqrt3)/2,(-1+isqrt3)/2` Using this, we have:

`(x-(-3))(x-(-1-isqrt3)/2)(x-(-1+isqrt3)/2)`

`=>(x+3)(x-(-1/2-(isqrt3)/2))(x-(-1/2+(isqrt3)/2))`

`=>(x+3)(x+1/2+(isqrt3)/2)(x+1/2-(isqrt3)/2)`