Question 669024
My solution mmay not be the expected way to go about it, but here it is.
The four answers feature four different quartic polynomials.
I'll call them:
{{{A(x)= x^4 + 5x^3 + 5x^2 + 5x + 4}}}
{{{B(x)= x^4 - 5x^3 - 5x^2 - 5x - 4}}}
{{{C(x)= -x^4 + 5x^3 + 5x^2 + 5x + 4}}}
{{{D(x)= x^4 + 5x^3 + 5x^2 + 5x - 5}}}
For {{{x=-1}}}, their values are:
{{{A (-1) = (-1)^4 + 5 (-1)^3 + 5 (-1)^2 + 5(-1) + 4=1-5+5-5+4=0}}}
{{{B(-1)= (-1)^4 - 5 (-1)^3 - 5 (-1)^2 - 5(-1) - 4=1+5-5+5-4=2}}}
{{{C(x)= -(-1)^4 + 5 (-1)^3 + 5 (-1)^2 + 5(-1) + 4=-1-5+5-5+4=-2}}}
{{{D(-1)= (-1)^4 + 5 (-1)^3 + 5 (-1)^2 + 5(-1) - 5=1-5+5-5-5=-9}}}
Only {{{a(x) has {{{x=-1}}} for a zero.
Does it have {{{x=-4}}} for a zero?
Does it have any other zeros?
You would expect it to have {{{x=-4}}} for a zero, and to have no other real zeros.
I would chose answer A and be done.
However, I can answer those two questions by factoring the polynomial.
{{{A(x)= x^4 + 5x^3 + 5x^2 + 5x + 4=(x^4+x^3) + (4x^3+4x^2) + (x^2+x) + (4x+ 4)=(x+1)(x^3+4x^2+x+4)}}}
{{{A(x)=(x+1)(x^3+4x^2+x+4)=(x+1)((x^3+4x^2)+(x+4))=(x+1)((x^2+1)(x+4))=(x+1)(x+4)(x^2+1)}}}
That is the complete factorization.
Because {{{x^2+1=0}}} has no real solutions, teh only zeros are the values of {{{x}}} that make the other two factors zero.
{{{x+1=0}}} <--> {{{highlight(x=-1)}}} and {{{x+4=0}}} <--> {{{highlight(x=-4)}}}