Question 30879
{{{ 2x^2 + 16x + 32 }}}
{{{ 2(x^2 + 8x + 16) }}}
{{{ 2(x+4)(x+4) }}}
{{{ 2(x+4)^2 }}}


{{{ x^3 + 3x^2 + x + 3 }}}
You need to find a factor of this ie a value of x that makes the whole thing zero. Since all the terms are positive, then x must be a negative value, to make the {{{x^3}}} and x negative.


As to the value of x, well this is trial and error to find, but looking at the equation, the value -3 does jump out. Put x=-3 into the expression and you do get zero. So, x+3 is a factor.


You now need to divide this into the cubic expression and then factorise the resulting quadratic, if possible. Having done this for you, the final answer is {{{(x+3)(x^2+1)}}}. So try it.


Actually, there is a simpler method for this cubic: {{{ x^3 + 3x^2 + x + 3 }}} written as {{{ x^3 + x + 3x^2 + 3 }}}. Look the 2 sets of 2 terms and factorise each of those:


{{{ (x^3 + x) + (3x^2 + 3) }}}
{{{ x(x^2 + 1) + 3(x^2 + 1) }}}


You can see that both halves have {{{x^2+1}}}, so we can factorise again, as {{{ (x^2 + 1)(x + 3) }}}


Third example looks to be like the first.


jon.