Factor by any method. See Examples 1–7. q2+6q+9−$$p2q^2$+6q+9-p^2$$

Ch. R - Review of Basic Concepts

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Lial 13th Edition

Ch. R - Review of Basic Concepts

문제 106

1장, 문제 106

Factor by any method. See Examples 1–7. $q^{2} + 6 q + 9 - p^{2}$

검증된 단계별 안내

Recognize that the expression $q^2 + 6q + 9 - p^2$ can be viewed as a difference of two parts: $(q^2 + 6q + 9)$ and $p^2$.

Notice that $q^2 + 6q + 9$ is a perfect square trinomial because it matches the form $a^2 + 2ab + b^2$, where $a = q$ and $b = 3$, so it factors as $(q + 3)^2$.

Rewrite the original expression as $(q + 3)^2 - p^2$ to clearly see it as a difference of squares.

Apply the difference of squares formula, which states that $A^2 - B^2 = (A - B)(A + B)$, where $A = (q + 3)$ and $B = p$.

Write the factored form as $((q + 3) - p)((q + 3) + p)$.

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주요 개념

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Factoring Quadratic Trinomials

Factoring quadratic trinomials involves expressing a quadratic expression like q² + 6q + 9 as a product of two binomials. Recognizing perfect square trinomials, where the first and last terms are perfect squares and the middle term is twice the product of their roots, helps simplify the process.

Difference of Squares

The difference of squares is a factoring technique used when an expression is in the form a² - b². It factors into (a - b)(a + b). Identifying this pattern allows for quick simplification of expressions like (q + 3)² - p².

Combining Factoring Methods

Some expressions require multiple factoring methods applied sequentially. For example, first factoring a perfect square trinomial, then applying the difference of squares formula. Understanding how to combine these techniques is essential for fully factoring complex expressions.

Factor by any method. See Examples 1–7. q2+6q+9−p2q^2+6q+9-p^2