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Ede4ka [16]
3 years ago
5

Simplify 3(x-4)+8(x+2) A) 11x-4 B)11x+4 C)11x-2

Mathematics
1 answer:
swat323 years ago
7 0
It is depending upon multiplication of values to variable and values of numbers again. Here the answer is Option B) 11x + 4. Here is my process on this equating expression.

For this we have to apply the law of distribution or the distributive law, in this we distribute the parentheses into the brackets to further expand them, we further eliminate the bracketed groups and perform the calculations. So, it is stating that, for any number, variable or a quantity if multiplied into a bracketed expression yields two products of multiplication of two quantities; one being the first quantity outside the brackets and two separately distinguished quantities inside the brackets. They are furthermore subtracted by the first quantity to that of the second quantity to obtain and achieve the distributive law.

If I were to express it through mathematical LaTeX Expressions, it will be represented as the first quantity as a variable of "a" where it can exhibit a value or a variable or a term or rule, etc. and then inside the brackets represented as "b" and "c".

Performing the law of distribution of values, variables, etc. It is demonstrated as:

\boxed{\mathbf{a \times (b - c) = a \times b - a \times c}}

We are given the values of the variables as a separate context and addition of two expressions. So, for both of them:

\mathbf{First \: \: Expression: \: 3 (x - 4)}

\mathbf{Second \: \: Expression: \: 8 (x + 2)}

Apply the following distribution law to expand the values onto variables and numbered values inside the brackets.

So, the variable "a", "b" and "c"; For First Expression is: a = 3, b = x and c = 4.

And, Similarly, for the variable of "a", "b" and "c", For Second Expression is: a = 8, b = x and c = 2.

Therefore, According to our distributive law of expanding the components will directly yield us:

\mathbf{3 \times x - 3 \times 4 + 8 \times x + 8 \times 2}

\mathbf{3x - 12 + 8x + 16}

Regroup the terms for a better visualisation and a clearer understanding for similarities between the terms; it makes it more easier to proceed further. So:

\mathbf{3x + 8x - 12 + 16}

All we did, was to separate the variable attached numbered values and shift it on one side; And, separating or shifting the values without variables to the other side.

\mathbf{11x - 12 + 16}

\boxed{\huge{\mathbf{Required \: \: Answer: \: 11x + 4}}}

Hope it helps.
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How does it factor out as that?
eimsori [14]
To factor out you have to think what multiples to AC and adds to B.

Ax^2+Bx+C

So... for this problem AxC=1x-24 or -24

B is -2.

So what two numbers multiply to -24: -3x8, -8x3, -4x6, -6x4, -2x12, -12x2.

Out of these, which adds to -2: -6+4=-2.

So the factors are (d-6)(d+4)

OR the longer way, which you really only use if A is not equal to 1.

Use the terms above and then rewrite the equation with two middle terms: d^2+4d-6d-24

Group the terms by using addition: (d^2+4d)+(6d-24)

Find what they have in common and factor it out. For the first, it's d. They both have d. So: d(d+4)
To check this, distribute the d. It should equal the first set lf parenthesis.

For the second, they have a number in common. 6 is a multiple of 24 so you can take that out: -6(d+4)

If the terms inside the parenthesis are the same, that's good. It means we can pair the insides and the outsides together to form the factors.

The two terms outside the parenthesis: d, -6 group together and become (d-6)

The inside terms stay the same: (d+4)

(d-6)(d+4)

Again, this is the longer way and no necessary for a problem like this. But if it was 2d^2, then this would be perfecf.




6 0
3 years ago
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