Math Problem Statement

Factor the following completely: a. xy + 2x + y + 2 b. x^3 + x^2 + x + 1 c. 2x^2 + 7x + 5 d. -5x^2 + x + 6 e. x^4 + 3x^2 - 4x^2 - 12 f. (3x + 2)^2 + 9x + 6 g. x^2 - 5x + 6

Solution

Let’s work through the factoring of each expression in the image:

a. xy+2x+y+2xy + 2x + y + 2

We can factor this by grouping:

(xy+2x)+(y+2)=x(y+2)+1(y+2)=(x+1)(y+2)(xy + 2x) + (y + 2) = x(y + 2) + 1(y + 2) = (x + 1)(y + 2)

So, the factored form is:

(x+1)(y+2)(x + 1)(y + 2)


b. x3+x2+x+1x^3 + x^2 + x + 1

We can use grouping again here:

(x3+x2)+(x+1)=x2(x+1)+1(x+1)=(x2+1)(x+1)(x^3 + x^2) + (x + 1) = x^2(x + 1) + 1(x + 1) = (x^2 + 1)(x + 1)

So, the factored form is:

(x2+1)(x+1)(x^2 + 1)(x + 1)


c. 2x2+7x+52x^2 + 7x + 5

We use the AC method for factoring quadratic expressions. Find two numbers that multiply to 2×5=102 \times 5 = 10 and add to 77, which are 55 and 22:

2x2+5x+2x+5=(2x2+5x)+(2x+5)2x^2 + 5x + 2x + 5 = (2x^2 + 5x) + (2x + 5)

Factor each group:

x(2x+5)+1(2x+5)=(2x+5)(x+1)x(2x + 5) + 1(2x + 5) = (2x + 5)(x + 1)

So, the factored form is:

(2x+5)(x+1)(2x + 5)(x + 1)


d. 5x2+x+6-5x^2 + x + 6

First, factor out 1-1:

(5x2x6)-(5x^2 - x - 6)

Now, factor 5x2x65x^2 - x - 6. We need two numbers that multiply to 5×(6)=305 \times (-6) = -30 and add to 1-1, which are 55 and 6-6:

5x2+5x6x6=(5x2+5x)+(6x6)5x^2 + 5x - 6x - 6 = (5x^2 + 5x) + (-6x - 6)

Factor each group:

5x(x+1)6(x+1)=(x+1)(5x6)5x(x + 1) - 6(x + 1) = (x + 1)(5x - 6)

So, the factored form is:

(x+1)(5x6)-(x + 1)(5x - 6)


e. x4+3x24x212x^4 + 3x^2 - 4x^2 - 12

Combine like terms:

x4x212x^4 - x^2 - 12

Let y=x2y = x^2, so the expression becomes:

y2y12y^2 - y - 12

Factor this as a quadratic:

(y4)(y+3)(y - 4)(y + 3)

Substitute back y=x2y = x^2:

(x24)(x2+3)(x^2 - 4)(x^2 + 3)

Factor further:

(x2)(x+2)(x2+3)(x - 2)(x + 2)(x^2 + 3)

So, the factored form is:

(x2)(x+2)(x2+3)(x - 2)(x + 2)(x^2 + 3)


f. (3x+2)2+9x+6(3x + 2)^2 + 9x + 6

Expand the square:

(3x+2)2=9x2+12x+4(3x + 2)^2 = 9x^2 + 12x + 4

So the expression becomes:

9x2+12x+4+9x+6=9x2+21x+109x^2 + 12x + 4 + 9x + 6 = 9x^2 + 21x + 10

Factor using the AC method. We need two numbers that multiply to 9×10=909 \times 10 = 90 and add to 2121, which are 1515 and 66:

9x2+15x+6x+10=(9x2+15x)+(6x+10)9x^2 + 15x + 6x + 10 = (9x^2 + 15x) + (6x + 10)

Factor each group:

3x(3x+5)+2(3x+5)=(3x+5)(3x+2)3x(3x + 5) + 2(3x + 5) = (3x + 5)(3x + 2)

So, the factored form is:

(3x+5)(3x+2)(3x + 5)(3x + 2)


g. x25x+6x^2 - 5x + 6

We need two numbers that multiply to 66 and add to 5-5, which are 2-2 and 3-3:

x25x+6=(x2)(x3)x^2 - 5x + 6 = (x - 2)(x - 3)

So, the factored form is:

(x2)(x3)(x - 2)(x - 3)


Would you like more details or explanations on any of these solutions?

Here are five related questions you might find interesting:

  1. How do you factor higher-degree polynomials using synthetic division?
  2. What are the different methods to factor trinomials?
  3. How do you use the difference of squares in factoring?
  4. Can you explain how to factor by grouping in more detail?
  5. How do you apply the quadratic formula to factor?

Tip: Always look for common factors first before using advanced factoring techniques like grouping or the quadratic method.

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Math Problem Analysis

Mathematical Concepts

Factoring
Polynomial Equations
Grouping Method
Quadratic Equations
Difference of Squares

Formulas

Quadratic equation formula: ax^2 + bx + c = 0
Factoring by grouping
Difference of squares: a^2 - b^2 = (a - b)(a + b)

Theorems

Factor Theorem
Difference of Squares Theorem

Suitable Grade Level

Grades 8-10