ac+cd-ab-bd = (a + d)(c – b)

because when you use the FOIL methods and multiply (a+d)(c-b) out… you get…

first terms…. (a)(c) = ac

outer terms… (a)(-b) = -ab

inner terms… (d)(c) = cd

last terms…. (d)(-b) = -bd

Now add up all of those terms…. and you get…

ac – ab + cd – bd…. which is the same as what you were given….ac+cd-ab-bd…. just rearranged…. see that?

PROBLEM 2:

The last exponent is missing… but if you put a space between “^3” and the “+”…. and also put a space between the “+” and the (x-3)^3…. then YA! won’t think that the line is too long and chop it off….. for you problem… I’m just going to assume that the last exponent is “4”….

So this is how you factor….

(x-3)^2(2x+1)^3 + (x-3)^3(2x+1)^4

** Hint: How many (x-3)’s can you take from both terms… and how many (2x+1)’s can you take from both terms…. You can take two (x-3)’s and three (2x+1)’s from both terms… like this…

_____ …..____

……….v 2………..v 3

(x-3)^2 (2x+1)^3 [1+ (x-3)(2x+1)]

(x-3)^2 (2x+1)^3 [1+ (2x^2 -6x + x -3)]

(x-3)^2 (2x+1)^3 [2x^2 -5x -2]

combine all similar terms.

(ac-ab) + (cd – bd)

a(c-b) + d( c- b)

groupings.

answer : (a+d) ( c-b)

to check. apply foil method.

ac – ab + dc – bd