To factor the quadratic expression \(y = 6x^2 + 13x + 5\), we need to look for two numbers that multiply to \(a \cdot c\) (where \(a\) is the coefficient of \(x^2\) and \(c\) is the constant term) and add up to \(b\) (the coefficient of \(x\)).
- Here, \(a = 6\), \(b = 13\), and \(c = 5\).
- We need to multiply \(a\) and \(c\): \(6 \cdot 5 = 30\).
- Now we need two numbers that multiply to \(30\) and add up to \(13\). The numbers \(3\) and \(10\) fit because \(3 \cdot 10 = 30\) and \(3 + 10 = 13\).
Next, we can split the middle term using these two numbers:
\[ y = 6x^2 + 3x + 10x + 5 \]
Now group the terms:
\[ y = (6x^2 + 3x) + (10x + 5) \]
Factor out the common factors in each group:
\[ y = 3x(2x + 1) + 5(2x + 1) \]
Now, factor out the common binomial factor \((2x + 1)\):
\[ y = (2x + 1)(3x + 5) \]
So the factored form of \(y = 6x^2 + 13x + 5\) is:
\[ y = (2x + 1)(3x + 5) \]
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