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$=\frac{3x^{2}+7x + 4}{2x^{2}-11x + 5}$

Question

$=\frac{3x^{2}+7x + 4}{2x^{2}-11x + 5}$

Explanation:

Step1: Factor the numerator

Factor $3x^{2}+7x + 4$. We need two numbers that multiply to $3\times4 = 12$ and add up to 7. The numbers are 3 and 4. So, $3x^{2}+7x + 4=3x^{2}+3x+4x + 4=3x(x + 1)+4(x + 1)=(3x + 4)(x+1)$.

Step2: Factor the denominator

Factor $2x^{2}-11x + 5$. We need two numbers that multiply to $2\times5=10$ and add up to - 11. The numbers are -10 and -1. So, $2x^{2}-11x + 5=2x^{2}-10x-x + 5=2x(x - 5)-(x - 5)=(2x - 1)(x - 5)$.

Step3: Write the factored - form of the fraction

The original fraction $\frac{3x^{2}+7x + 4}{2x^{2}-11x + 5}=\frac{(3x + 4)(x + 1)}{(2x - 1)(x - 5)}$

Answer:

$\frac{(3x + 4)(x + 1)}{(2x - 1)(x - 5)}$