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Factorize the following algebraic expressions: 2.2 (xy + 5y) + (2x + 10) 2.3 Factor these values out of each group and then write down the equivalent algebraic expression. 2.4 What is the common facto

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4 parts
2.3

y(x + 5) + 2(x + 5)

2.4

(x + 5)

2.5

(x + 5)(y + 2)

2.6

(x + 2)(3x + 4)

Step-by-step solution

7 steps

Here are the solutions to the factorization problems:

Part 2.2

If we group the first two terms and the last two terms as follows: (xy+5y)+(2x+10)(xy + 5y) + (2x + 10)

what do you notice about each group?
Each group has a common factor. In Group 1, yy is common. In Group 2, 2 is common.

Part 2.3y(x + 5) + 2(x + 5)

Factor these values out of each group and then write down the equivalent algebraic expression.

  1. Factor Group 1.

    xy+5y=y(x+5)xy + 5y = y(x + 5)
  2. Factor Group 2.

    2x+10=2(x+5)2x + 10 = 2(x + 5)
  3. Write the equivalent algebraic expression.

    y(x+5)+2(x+5)y(x + 5) + 2(x + 5)

    The equivalent algebraic expression is y(x+5)+2(x+5)\boxed{y(x + 5) + 2(x + 5)}.

Part 2.4(x + 5)

What is the common factor in the two terms?

The common factor in the two terms y(x+5)y(x + 5) and 2(x+5)2(x + 5) is (x+5)\boxed{(x + 5)}.

Part 2.5(x + 5)(y + 2)

Use the distributive property to factor out this common factor and then express the polynomial as a product of two binomials.

  1. Factor out the common binomial factor (x+5)(x + 5).

    y(x+5)+2(x+5)=(x+5)(y+2)y(x + 5) + 2(x + 5) = (x + 5)(y + 2)

    The polynomial as a product of two binomials is (x+5)(y+2)\boxed{(x + 5)(y + 2)}.

Part 2.6(x + 2)(3x + 4)

Joseph wants to factorize the following algebraic expression: 3x2+6x+4x+83x^2 + 6x + 4x + 8

Provide Joseph with a three-step guide on how to factorize the expression.

  1. Group the terms into two pairs.

    (3x2+6x)+(4x+8)(3x^2 + 6x) + (4x + 8)
  2. Factor out the greatest common factor (GCF) from each group.

    From the first group, 3x2+6x3x^2 + 6x, the GCF is 3x3x, so it becomes 3x(x+2)3x(x + 2).
    From the second group, 4x+84x + 8, the GCF is 4, so it becomes 4(x+2)4(x + 2).
    The expression is now 3x(x+2)+4(x+2)3x(x + 2) + 4(x + 2).

  3. Factor out the common binomial factor.

    The common binomial factor is (x+2)(x + 2).

    (x+2)(3x+4)(x + 2)(3x + 4)

    The factored expression is (x+2)(3x+4)\boxed{(x + 2)(3x + 4)}.

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