Question 726930
There must be some quadratic polynomial {{{ax^2+bx+c}}} that squares to yield
{{{x^4 - 14x^3 + 71x^2 + px + q }}} .
Since the leading coefficient in {{{x^4 - 14x^3 + 71x^2 + px + q }}} is {{{1}}} ,
we know that {{{a=1}}} , but we still have to find {{{b}}} and {{{c}}}.
By painfully multiplying we find that
{{{(x^2+bx+c)(x^2+bx+c)= x^4 +2bx^3 +(b^2+2c)x^2 +2bcx + c^2}}}
If the polynomials {{{x^4 - 14x^3 + 71x^2 + px + q }}} and {{{x^4 +2bx^3 +(b^2+2c)x^2 +2bcx + c^2}}}
are the same, then the coefficients are the same
{{{2b=-14}}}
{{{b^2+2c=71}}}
{{{p=2bc}}} and
{{{q=c^2}}}
The first two equations let us find {{{b}}} and {{{c}}} .
We can use the last two to calculate {{{p}}} and {{{q}}} from {{{b}}} and {{{c}}}.
{{{2b=-14}}} --> {{{b=-7}}}
{{{b^2+2c=71}}} --> {{{(-7)^2+2c=71}}} --> {{{49+2c=71}}} --> {{{2c=22}}} --> {{{c=11}}}
{{{p=2bc}}} --> {{{p=2(-7)(11)}}} --> {{{highlight(p=-154)}}}
{{{q=c^2}}} --> {{{q=11^2}}} --> {{{highlight(q=121)}}}