Question

1)3а²+10а+3=(3а+9)(а+⅓)2)(а+1)(а²+5а+6)=(а²+3а+2)(а+3)​

Answer 1

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Answer 2

It seems like you have two expressions that can be factored:

1. $3a^2 + 10a + 3 = (3a + 9)(a + 1)$

To factor this expression, you can use the distributive property:

$3a^2 + 10a + 3 = 3a^2 + 9a + a + 3$

Now, you can factor by grouping:

$3a^2 + 9a + a + 3 = 3a(a + 3) + 1(a + 3)$

Now, you can factor out the common factor of $a + 3$:

$3a(a + 3) + 1(a + 3) = (3a + 1)(a + 3)$

So, the factorization of $3a^2 + 10a + 3$ is $(3a + 1)(a + 3)$.

2. $a^2 + 5a + 6 = (a^2 + 3a + 2)(a + 3)$

To factor this expression, you can look for two numbers that multiply to the constant term (6) and add up to the coefficient of the middle term (5). In this case, those numbers are 2 and 3:

$a^2 + 5a + 6 = a^2 + 2a + 3a + 6$

Now, you can factor by grouping:

$a^2 + 2a + 3a + 6 = (a^2 + 2a) + (3a + 6)$

Factor out the common factors from each group:

$a^2 + 2a = a(a + 2)$

$3a + 6 = 3(a + 2)$

Now, you have:

$(a^2 + 2a) + (3a + 6) = a(a + 2) + 3(a + 2)$

Factor out the common factor of $a + 2$:

$a(a + 2) + 3(a + 2) = (a + 2)(a + 3)$

So, the factorization of $a^2 + 5a + 6$ is $(a + 2)(a + 3)$.