Which expression is equal to (x-3)(2x^2-x+3)

2x^3 - 7x^2 + 6x - 9 which is choice D

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Work Shown:

One way is to use the distribution rule two times

(x-3)(2x^2-x+3) = y(2x^2-x+3) ........... replace (x-3) with y

(x-3)(2x^2-x+3) = y(2x^2)+y(-x)+y(3) .... distribution rule

(x-3)(2x^2-x+3) = 2x^2*(y) - x(y) + 3(y)

(x-3)(2x^2-x+3) = 2x^2*(x-3) - x(x-3) + 3(x-3) .... replace y with x-3

(x-3)(2x^2-x+3) = 2x^3-6x^2 - x^2 + 3x + 3x - 9 ... distribution rule

(x-3)(2x^2-x+3) = 2x^3 - 7x^2 + 6x - 9

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Or we can use the box method (see the attached image below). This is a visual way to organize the terms. You'll probably notice that the box method is basically the distribution rule. Each row is one distribution being applied. In row 1, we have x distributed to (2x^2-x+3). In row 2, we have -3 distributed to (2x^2-x+3). I color coded the table cells to highlight the like terms. Those like terms combine to -6x^2-x^2 = -7x^2 and 3x+3x = 6x as shown in the steps for the distribution above.

Each interior cell in the box is found by multiplying the corresponding outer terms. For example, in the first row, first column we have x^3 which is the result of multiplying the outer x and x^2 terms.

if this is really you s much. i only have 3 points left and i have 2 questions to ask

geometric series converge the common ratio r has magnitude less than one.

which is equivalent to

for the example we need

|2-x| < 1

-1 < 2-x   and 2-x < 1

x < 3 and x > 1

we can write that

answer: 1 < x < 3

1a. common ratio r=250/25=10 > 1     diverges

1b.   r=50/100=1/2 converges

when these converge they converge to

where a is the first term.

answer: 200

1c. r=-1   so it's not the case that |r| < 1.     diverges

2a. x is the common ratio so we converge when |x|< 1

2b.   r=-10x / -2 = 5x.

we converge when

|5x| < 1

|x| < 1/5

2c.   r=3-2x

-1 < 3-2x and 3-2x < 1

-4 < -2x and 2 < 2x

x < 2 and 1 < x

1 < x < 2

2d. [/tex] r=-48x^3/12 = 4x^3[/tex]

|4x^3| < 1

|x^3| < \frac 1 4

|x| < 4^{-1/3}