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

2 Answers
George C.
Mar 31, 2018

#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)#

Parabola
Mar 31, 2018

#(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:
enter image source here

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)#

Related questions
Impact of this question
1415 views around the world
You can reuse this answer
Creative Commons License