Calculus

4 (x+3)2 - 5(x+3)

First of all, there's no equation here, so it's not possible to solve it and find

a value for 'x'. Therefore, our next question has to be: What exactly do you

need to "do" with it ? We're suspecting that it's an exercise in elementary

algebra, and the assignment is to re-write this expression in a simpler form,

just to give you some practice in handling algebraic expressions. Understand

that if we do this, the result will not be an "answer" or a "solution" to anything

except the teacher's assignment, which was to write the same thing in a

different way.

We can see two different ways to go about this exercise. They should both

lead to the same thing. Since the purpose of the whole thing is to practice

these manipulations, we'll demonstrate both ways. That will give us the practice,

and we hope that you'll then try and go through it so that you also get the practice.

4 (x+3)2 - 5(x+3)

Before we begin the first method, let's make the expression a bit simpler to

handle. Notice that (x+3) appears in two different places. To cut down slightly

on the amount if writing required, let's call that quantity 'Q' for only a few seconds.

Now we have

4 Q2 - 5Q. This can be factored: Q (4Q - 5), and we can now replace (x+3)

wherever we see 'Q':

(x+3) [ 4(x+3) - 5 ]

Inside the brackets, eliminate the parentheses: (x+3) [ 4x + 12 - 5 ] = (x+3) [ 4x + 7 ].

Now multiply the (x+3) on the left by the [4x+7] on the right:

4x2 + 7x + 12x + 21 = 4x2 + 19x + 21 and that's the result of our first method.

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Now here's our second method:

4 (x+3)2 - 5(x+3)

Square the first (x+3) as indicated, and put its square inside the first parentheses:

4 (x2+ 6x + 9) - 5(x+3)

Eliminate the first parentheses:

4x2+ 24x + 36 - 5(x+3)

Eliminate the second parentheses:

4x2 + 24x + 36 - 5x - 15

Combine the 'x' terms:

4x2 + 19x + 36 - 15

Combine the numerical terms:

4x2 + 19x + 21 and notice that this is the same as the result of the other method.