QUESTION IMAGE
Question
- (6x² - 13x + 5) ÷ (2x - 1)
no parenthesis
Step1: Factor the numerator
We factor the quadratic expression \(6x^2 - 13x + 5\). We need two numbers that multiply to \(6\times5 = 30\) and add up to \(-13\). The numbers are \(-10\) and \(-3\). So we rewrite the middle term:
$$
LATEXBLOCK0
$$
Step2: Divide the factored form by the denominator
Now we have \(\frac{(2x - 1)(3x - 5)}{2x - 1}\). Since \(2x-1
eq0\) (to avoid division by zero), we can cancel out the common factor \(2x - 1\) from the numerator and the denominator.
$$
\frac{(2x - 1)(3x - 5)}{2x - 1}=3x - 5
$$
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\(3x - 5\)