what is the quotient of $(x^{3}+3x^{2}-4x - 12)div(x^{2}+5x + 6)$?\n$x + 2$\n$x - 2$\n$x - 8$\n$x + 8$
what is the quotient of $(x^{3}+3x^{2}-4x - 12)div(x^{2}+5x + 6)$?\n$x + 2$\n$x - 2$\n$x - 8$\n$x + 8$
Answer
Answer:
B. $x - 2$
Explanation:
Step1: Factor the numerator
Factor $x^{3}+3x^{2}-4x - 12$ by grouping. Group the terms: $(x^{3}+3x^{2})+(-4x - 12)$. Factor out the greatest - common factor from each group: $x^{2}(x + 3)-4(x + 3)=(x + 3)(x^{2}-4)$. Using the difference - of - squares formula $a^{2}-b^{2}=(a + b)(a - b)$, we can further factor $x^{2}-4$ as $(x + 2)(x - 2)$. So, $x^{3}+3x^{2}-4x - 12=(x + 3)(x + 2)(x - 2)$.
Step2: Factor the denominator
Factor $x^{2}+5x + 6$. We need to find two numbers that multiply to $6$ and add up to $5$. The numbers are $2$ and $3$. So, $x^{2}+5x + 6=(x + 2)(x+3)$.
Step3: Simplify the rational expression
The original expression $\frac{x^{3}+3x^{2}-4x - 12}{x^{2}+5x + 6}=\frac{(x + 3)(x + 2)(x - 2)}{(x + 2)(x + 3)}$. Cancel out the common factors $(x + 2)$ and $(x + 3)$ in the numerator and the denominator. We get $x - 2$.