Question 148487
The expression "the product of 3 less than twice x and 2 more than the quanity 3 times x" translates to {{{(2x-3)(3x+2)}}}



So we need to expand {{{(2x-3)(3x+2)}}}




{{{(2x-3)(3x+2)}}} Start with the given expression.



Now let's FOIL the expression.



Remember, when you FOIL an expression, you follow this procedure:



{{{(highlight(2x)-3)(highlight(3x)+2)}}} Multiply the <font color="red">F</font>irst terms:{{{(2*x)*(3*x)=6*x^2}}}.



{{{(highlight(2x)-3)(3x+highlight(2))}}} Multiply the <font color="red">O</font>uter terms:{{{(2*x)*(2)=4*x}}}.



{{{(2x+highlight(-3))(highlight(3x)+2)}}} Multiply the <font color="red">I</font>nner terms:{{{(-3)*(3*x)=-9*x}}}.



{{{(2x+highlight(-3))(3x+highlight(2))}}} Multiply the <font color="red">L</font>ast terms:{{{(-3)*(2)=-6}}}.



{{{6*x^2+4*x-9*x-6}}} Now collect every term to make a single expression.



{{{6*x^2-5*x-6}}} Now combine like terms.



So {{{(2x-3)(3x+2)}}} FOILS to {{{6*x^2-5*x-6}}}.



In other words, {{{(2x-3)(3x+2)=6*x^2-5*x-6}}}.



So this means that the answer is choice d)