Math Problem Statement

answer with complete solution

Factor each of the polynomials completely, if possible. If the polynomial cannot be factored, state that it is prime.

Solution

Let's solve these polynomial factorization problems one by one.

36) x26xy+9y2x^2 - 6xy + 9y^2

This is a quadratic trinomial. It resembles a perfect square trinomial:

x26xy+9y2=(x3y)2x^2 - 6xy + 9y^2 = (x - 3y)^2

Thus, the factored form is:

(x3y)2(x - 3y)^2

37) 9x2a2x39x2+a29x^2 - a^2x^3 - 9x^2 + a^2

First, let's simplify:

9x29x2+a2a2x3=a2(1x3)9x^2 - 9x^2 + a^2 - a^2x^3 = a^2(1 - x^3)

Factoring 1x31 - x^3 using the difference of cubes:

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

Thus, the factored form of the entire expression is:

a2(1x)(1+x+x2)a^2(1 - x)(1 + x + x^2)

38) 3xy5rx10rs+6sy3xy - 5rx - 10rs + 6sy

Group the terms to factor by grouping:

(3xy+6sy)(5rx+10rs)(3xy + 6sy) - (5rx + 10rs)

Factor out common terms in each group:

3y(x+2s)5r(x+2s)3y(x + 2s) - 5r(x + 2s)

Now factor the common binomial x+2sx + 2s:

(3y5r)(x+2s)(3y - 5r)(x + 2s)

39) a2b225c2a^2b^2 - 25c^2

This is a difference of squares:

a2b225c2=(ab5c)(ab+5c)a^2b^2 - 25c^2 = (ab - 5c)(ab + 5c)

40) 6x2+13xy5y26x^2 + 13xy - 5y^2

We need to factor this trinomial. We look for two numbers that multiply to 6×5=306 \times -5 = -30 and add to 1313. These numbers are 1515 and 2-2. Thus, rewrite the middle term:

6x2+15xy2xy5y26x^2 + 15xy - 2xy - 5y^2

Group terms:

(6x2+15xy)(2xy+5y2)(6x^2 + 15xy) - (2xy + 5y^2)

Factor each group:

3x(2x+5y)y(2x+5y)3x(2x + 5y) - y(2x + 5y)

Now factor the common binomial 2x+5y2x + 5y:

(3xy)(2x+5y)(3x - y)(2x + 5y)

Summary of Solutions:

  1. x26xy+9y2=(x3y)2x^2 - 6xy + 9y^2 = (x - 3y)^2
  2. 9x2a2x39x2+a2=a2(1x)(1+x+x2)9x^2 - a^2x^3 - 9x^2 + a^2 = a^2(1 - x)(1 + x + x^2)
  3. 3xy5rx10rs+6sy=(3y5r)(x+2s)3xy - 5rx - 10rs + 6sy = (3y - 5r)(x + 2s)
  4. a2b225c2=(ab5c)(ab+5c)a^2b^2 - 25c^2 = (ab - 5c)(ab + 5c)
  5. 6x2+13xy5y2=(3xy)(2x+5y)6x^2 + 13xy - 5y^2 = (3x - y)(2x + 5y)

Would you like further details on any of these steps?

5 Related Questions:

  1. How can you recognize a perfect square trinomial?
  2. What is the difference between factoring by grouping and factoring a difference of squares?
  3. How do you factor higher-degree polynomials using special identities like the sum or difference of cubes?
  4. Why is factoring important in simplifying rational expressions?
  5. What is the general process for factoring trinomials of the form ax2+bx+cax^2 + bx + c?

Tip:

Always check your factored answer by expanding it back out to ensure it matches the original expression!

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Math Problem Analysis

Mathematical Concepts

Factoring Polynomials
Algebraic Expressions
Difference of Squares
Trinomials
Factoring by Grouping

Formulas

Difference of Squares: a^2 - b^2 = (a - b)(a + b)
Perfect Square Trinomial: a^2 - 2ab + b^2 = (a - b)^2

Theorems

Difference of Cubes
Factoring Trinomials

Suitable Grade Level

Grades 9-11