what is the correct factorization of $x^{2}+5x - 6$?\n$(x - 1)(x + 6)$\n$(x + 1)(x - 6)$\n$(x - 2)(x…
what is the correct factorization of $x^{2}+5x - 6$?\n$(x - 1)(x + 6)$\n$(x + 1)(x - 6)$\n$(x - 2)(x - 3)$\n$(x - 2)(x + 3)$
Answer
Explanation:
Step1: Recall factoring quadratic form
For a quadratic $ax^{2}+bx + c$ (here $a = 1$, $b=5$, $c=-6$), we need two numbers $m$ and $n$ such that $m\times n=ac=-6$ and $m + n=b = 5$.
Step2: Find the two numbers
The numbers are $-1$ and $6$ since $(-1)\times6=-6$ and $-1 + 6=5$.
Step3: Rewrite the middle - term
$x^{2}+5x - 6=x^{2}-x+6x - 6$.
Step4: Group the terms
$(x^{2}-x)+(6x - 6)=x(x - 1)+6(x - 1)$.
Step5: Factor out the common factor
$(x - 1)(x + 6)$.
Answer:
A. $(x - 1)(x + 6)$