```Question 324557
1.  {{{(x+3)^2-10(x+3)+24}}} Factor completely
{{{y^2-10y+24}}} Factor-look for 2 numbers that mult to 24 and add or subtract to get -10. They are -6 and -4.
{{{(y-6)(y-4)}}}
Now substitute (x+3) back in for y.
{{{(x+3-6)(x+3-4)}}}simplify
{{{(x-3)(x-1)}}}

2.  Write a quadratic equation and answer the following: the shortest leg of a right triangle is two less than the hypotenuse. The hypotenuse is one longer than the longest leg.

let shortest leg=a, let longer leg=b, let hypotenuse=c
{{{a=c-2}}}
{{{c=b+1}}}
substitute b+1 for c in first equation
{{{a=b+1-2}}}
{{{a=b-1}}} simplified
to summarize: {{{a=b-1}}},{{{b=b}}},{{{c=b+1}}}a, b, and c are all in terms of b.
{{{a^2+b^2=c^2}}} Pythagorean theorem
{{{(b-1)^2+b^2=(b+1)^2}}} Pyth thm with b-1 substituted for a and b+1 substituted for c.
{{{(b^2-2b+1)+b^2=(b^2+2b+1)}}} FOIL {{{(b-1)^2}}} and {{{(b+1)^2}}}
{{{2b^2-2b+1=b^2+2b+1}}} simplified
{{{b^2-2b+1=2b+1}}} b^2 subtracted from both sides
{{{b^2-4b=0}}} 2b and 1 subtracted from both sides
{{{b(b-4)=0}}} factor out b
{{{b=0andb-4=0}}} set each factor equal to zero and solve
{{{b=0andb=4) solve each equation.
{{{b=4}}}Since this is a real life geometry problem, b=0 does not make sense and is therefore excluded as an answer.
{{{a=4-1=3}}}and {{{c=4+1=5}}} 4 substituted in for b in a=b-1 and c=b+1 equations```