Question 71901
Factor:
(a) {{{y^2 + 16y + 64}}} Notice that the first and last terms are perfect squares. This suggests the possibility that the expression itself may be a perfect square. Since {{{sqrt(64) = 8}}}, you can try:
{{{y^2+16y+64 = (y+8)(y+8)}}}
Check using FOIL:
{{{(y+8)(y+8) = y^2+8y+8y+64}}} = {{{y^2+16y+64}}}
The factors are: {{{y+8}}} and {{{y+8}}}
(b) {{{by+7b-6y-42}}} Here, you use use "factor by grouping". Group the terms as folows:
{{{(by+7b) - (6y+42)}}} Notice the change of sign on the last term when the parentheses were added. Now, from each group, factor the common factors.
{{{b(y+7) - 6(y+7)}}} Now you can factor the common factor of (y+7)
{{{(y+7)(b-6)}}} These are the factors.
Solve:
(c) {{{x^2-x-20 = 0}}} This will factor. Notice that the last term has factors of 4 and 5. Also notice that 4-5 = -1 and this is the coefficient of the middle term. So we try:
{{{x^2-x-20 = (x+4)(x-5)}}}
{{{(x+4)(x-5) = 0}}} Apply the zero product principle:
{{{x+4 = 0}}} and/or {{{x-5 = 0}}}
If {{{x+4 = 0}}} then {{{x = -4}}}
If {{{x-5 = 0}}} then {{{x = 5}}}
The roots are:
{{{x = -4}}}
{{{x = 5}}}
(d) 
{{{9x^2 = 81}}} Notice that both sides are perfect squares.
{{{(3x)^2 = (9)^2}}} Take the square root of both sides.
{{{3x = sqrt(9^2)}}}
{{{3x = -9}}} or {{{3x = 9}}} Divide both sides by 3.
{{{x = -3}}} or {{{x = 3}}}