Factor the Expression: (x+8)(2y+16)+4(3x+24) Step by Step

Breakdown into factors by groups The given expression:

$(x+8)(2y+16)+4(3x+24)$This is done by extracting a common factor, both for the numerators and for the letters, in one of the parts of the given expression (the parts of the expression are separated by addition or subtraction operations between the multiplication terms) or both, in order to be able to distinguish a common multi-variable factor, and take it out of the parentheses, etc.

Let's refer separately to the numerators and letters, recalling that a common factor is a factor (multiplier) common to all the terms in the expression,

Let's start by examining the two parts of the expression separately, the first expression on the left:

$(x+8)(2y+16)$and the second expression on the left:

$4(3x+24)$Note that the expression in the parentheses in the second expression above and the expression in the parentheses in the first expression on the left in the original expression are proportional to each other (i.e. it is possible to get from one expression to the other by multiplication by some factor), this is because it is possible to take out a common factor out of the parentheses:

$4(3x+24) \\ 4\cdot3(x+8) \\ 12(x+8)$

$$We used the fact that the number 24 is a multiple of the number 3:

$24=8\cdot3$

Let's now go back to the original expression in the question and apply this knowledge:

$(x+8)(2y+16)+4(3x+24) \\ \downarrow\\ (x+8)(2y+16)+4\cdot3(x+8)\\ (x+8)(2y+16)+12(x+8)$Let's now use the distributive property and rearrange the expression we got again:

$(x+8)(2y+16)+12(x+8) \\ (2y+16)(x+8)+12(x+8)$Now we can note that in the expression we got in the last step there is a common factor which is multi-variable (i.e. includes more than one variable in the expression):

$x+8$This is because it is a multiple of both the first part of the expression on the left (the parentheses multiplier) and of the second part of the expression on the left:

$(2y+16)\underline{(x+8)}+12\underline{(x+8)}$Therefore, we can take out this expression in its entirety out of the parentheses as a common factor and break down the given expression in the usual way (i.e. using the answer to the question: "By what can we multiply the common factor (including its sign) in order to get each of the terms in the original expression (including their sign)?"):

$\textcolor{red}{ (2y+16)(x+8)}\textcolor{blue}{+12(x+8) } \\ \underline{(x+8)}\cdot\textcolor{red}{(2y+16)}+\underline{(x+8)}\cdot\textcolor{blue}{(+12)}\\ \downarrow\\ \underline{(x+8)}\big(\textcolor{red}{(2y+16)}\textcolor{blue}{+12}\big)$

In the expression above the operation is explained using colors and signs:

The common factor is highlighted using an underline, and the multipliers inside the parentheses correspond to the terms in the original expression using colors, note that we also referred to the sign, both of the common factor (in black) that we took out of the parentheses and to the signs of the terms in the original expression (in colors), there is no need to display this in steps as described above, you can (and should) jump directly to the broken down form in the last line, but definitely need to refer to the signs above, as in each term the sign is an integral part of it,

So we got the expression broken down into factors (by groups):

$(x+8)\big((2y+16)+12\big)$Let's continue and complete the breakdown while simplifying the expression in the right parentheses, note that the expression we got in the right parentheses in the multiplier of the parentheses we got in the last step, can be further broken down into factors by taking out the common factor the number: 2, this is because the number 28 is a multiple of the number 2:

$(x+8)(2y+28) \\ (x+8)\cdot2(y+14) \\ 2(x+8)(y+14)$In the last step we used the distributive property to rearrange the expression we got.

Let's summarize the steps of breaking down the expression (by groups), we got that:

$(x+8)(2y+16)+4(3x+24) \\ (x+8)(2y+16)+12(x+8) \\ (x+8)\big((2y+16)+12\big) \\ (x+8)(2y+28) \\ 2(x+8)(y+14)$We can be sure that this breakdown is correct easily by opening the parentheses using the extended distribution law and verifying that indeed the original expression we broke down is obtained term by term, this should be done while paying attention to the signs of the terms in the original expression and to the sign (given for selection always) of the common factor.

Therefore, the correct answer is answer c.