Match Binomial Expressions: (2x-y)(x+3) and Their Expanded Forms

Match the expressions (numbers) with the equivalent expressions (letters):

  1. (2xy)(x+3) (2x-y)(x+3)

  2. (y2x)(3x) (y-2x)(3-x)

  3. (2x+y)(x3) (2x+y)(x-3)

    a.2x26x+yx3y 2x^2-6x+yx-3y

    b.2x26xyx+3y 2x^2-6x-yx+3y

    c.2x2+6xyx3y 2x^2+6x-yx-3y

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Step-by-step video solution

Watch the teacher solve the problem with clear explanations
00:00 Open parentheses
00:04 Open parentheses properly, multiply each term by each term
00:16 Calculate the multiplications
00:21 And this is the simplification for 1, continue to 2
00:24 Open parentheses properly, multiply each term by each term
00:37 Calculate the multiplications
00:48 And this is the simplification for 2, continue to 3
00:51 Open parentheses properly, multiply each term by each term
00:57 Calculate the multiplications
01:04 And this is the solution to the question

Step-by-step written solution

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1

Understand the problem

Match the expressions (numbers) with the equivalent expressions (letters):

  1. (2xy)(x+3) (2x-y)(x+3)

  2. (y2x)(3x) (y-2x)(3-x)

  3. (2x+y)(x3) (2x+y)(x-3)

    a.2x26x+yx3y 2x^2-6x+yx-3y

    b.2x26xyx+3y 2x^2-6x-yx+3y

    c.2x2+6xyx3y 2x^2+6x-yx-3y

2

Step-by-step solution

Simplify the given expressions, open parentheses using the extended distributive property:

(a+b)(c+d)=ac+ad+bc+bd (\textcolor{red}{a}+\textcolor{blue}{b})(c+d)=\textcolor{red}{a}c+\textcolor{red}{a}d+\textcolor{blue}{b}c+\textcolor{blue}{b}d Keep in mind that in the formula form for the distributive property mentioned above, we assume by default that the operation between the terms inside the parentheses is an addition, therefore, of course, we will not forget that the sign of the term's coefficient is an inseparable part of it. Furthermore, we will apply the rules of sign multiplication and thus we can present any expression within parentheses, which is opened with the help of the previous formula, first, as an expression in which an addition operation takes place among all the terms (if necessary),

Then we will simplify each and every one of the expressions of the given problem, respecting the above, first opening the parentheses through the previously mentioned distributive property. Then we will use the substitution property in addition and multiplication before introducing like terms (if there are like terms in the expression obtained after opening the parentheses):

  1. (2xy)(x+3)(2x+(y))(x+3)2xx+2x3+(y)x+(y)32x2+6xyx3y (2x-y)(x+3) \\ \downarrow\\ \big(2x+(-y)\big)(x+3) \\ 2x\cdot x+2x\cdot 3+(-y)\cdot x+(-y)\cdot3\\ \boxed{2x^2+6x-yx-3y}\\

  2. (y2x)(3x)(y+(2x))(3+(x))y3+y(x)+(2x)3+(2x)(x)3yxy6x+2x2 (y-2x)(3-x) \\ \downarrow\\ \big(y+(-2x)\big)\big(3+(-x)\big) \\ y\cdot 3+y\cdot (-x)+(-2x)\cdot 3+(-2x)\cdot(-x)\\ \boxed{3y-xy-6x+2x^2}\\

  3. (2x+y)(x3)(2x+y)(x+(3))2xx+2x(3)+yx+y(3)2x26x+yx3y (2x+y)(x-3) \\ \downarrow\\ (2x+y)(x+(-3)) \\ 2x\cdot x+2x\cdot (-3)+y\cdot x+y\cdot(-3)\\ \boxed{2x^2-6x+yx-3y}\\ As you can notice, in all the expressions where we applied multiplication between the expressions in the previous parentheses, the result of the multiplication (obtained after applying the previously mentioned distributive property) produced an expression in which terms cannot be added, and this is because all the terms in the resulting expression are different from each other (remember that all like variables must be identical and in the same power),

    Now, let's use the substitution property in addition and multiplication to distinguish that:

    The simplified expression in 1 corresponds to the expression in option C,

    The simplified expression in 2 corresponds to the expression in option B,

    The simplified expression in 3 corresponds to the expression in option A,

Therefore, the correct answer (among the options offered) is option B.

3

Final Answer

1-b, 2-c, 3-a

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\( (3+20)\times(12+4)= \)

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