Either multiply the constant factor by the first linear
factor:
$$
\cl"tight"{\table , 2(x+4)(3x+2);
=, (2(x)+2(4))(3x+2);
=, (2x+8)(3x+2)}
$$
and then expand the result as you did in nonHomogQuestion:
$$
\cl"tight"{\table , \colspan 7 (2x+8)(3x+2);
=, \colspan 3 2x(3x+2), +, \colspan 3 8(3x+2);
=, 2x(3x), +, 2x(2), +, 8(3x), +, 8(2);
=, 6x^2, +, 4x, +, 24x, +, 16;
=, \colspan 7 \cl"highl"{6x^2 + 28x + 16}}
$$
| Or, multiply the two linear factors first:
$$
\cl"tight"{\table , 2(x+4)(3x+2);
=, 2(x(3x+2) + 4(3x+2));
=, 2(x(3x) + x(2) + 4(3x) + 4(2));
=, 2(3x^2 + 2x + 12x + 8);
=, 2(3x^2 + 14x + 8)}
$$
and then expand this as you did in the previous lesson:
$$
\cl"tight"{\table , 2(3x^2 + 14x + 8);
=, 2(3x^2) + 2(14x) + 2(8);
=, \cl"highl"{6x^2 + 28x + 16}}
$$
|