Question

Factor by Grouping.

Answer 1

Answer:

The ac product of $35x^2+41x+12$ is 420.

The factors of the ac product that add to 41 are 20 and 21.

$35x^2+41x+12 =$ $(7x+4)(5x+3)$

Step-by-step explanation:

1) The general form of a quadratic is $ax^2 + bx + c$. Hence, multiplying 35 by 12 gives you the product of ac, which is 420.

2) We need to find two numbers that multiply to 420 and add up to 41 simultaneously. If we pull out the factors of 420, two of them will be 20 and 21, which multiply to 420 as well as add up to 41.

3) Write $41x$ as a sum.

$35x^2+20x+21x+12$

Now, we can factor them out by grouping.

$35x^2+20x+21x+12\\5x(7x+4)+3(7x+4)$

Since, $7x +4$ is common in both of the factors, we only take one of the $7x+4$ along with $5x$ and $3$.

$(7x+4)(5x+3)$

Answer 2

Answer:

$\text{The ac product of }\: 35x^2+41x+12 \text{ is }\boxed{420}\:.$

$\text{The factors of the ac product that add to 41 are }\:\boxed{20} \: \text{ and }\: \boxed{21}\:.$

$35x^2+41x+12=\left( \: \boxed{5}\:x+\:\boxed{3}\:\right)\left(\: \boxed{7}\:x+\boxed{4}\:\right)$

Step-by-step explanation:

Given quadratic:

$35x^2+41x+12$

Factoring quadratics by grouping

To factor a quadratic in the form $ax^2+bx+c$,  find two numbers that multiply to ac and sum to b.

$\implies ac=35 \cdot 12 = 420$

$\implies b = 41$

Therefore, the two numbers that multiply to 420 and sum to 41 are:
20 and 21

Rewrite b as the sum of these two numbers:

$\implies 35x^2+20x+21x+12$

Factorize the first two terms and the last two terms separately:

$\implies 5x(7x+4)+3(7x+4)$

Factor out the common term $(7x+4)$ :

$\implies (5x+3)(7x+4)$

Conclusion

$\text{The ac product of }\: 35x^2+41x+12 \text{ is }\boxed{420}\:.$

$\text{The factors of the ac product that add to 41 are }\:\boxed{20} \: \text{ and }\: \boxed{21}\:.$

$35x^2+41x+12=\left( \: \boxed{5}\:x+\:\boxed{3}\:\right)\left(\: \boxed{7}\:x+\boxed{4}\:\right)$

Learn more here:

brainly.com/question/27929560