Answer:
[tex]\text{The ac product of }\: 35x^2+41x+12 \text{ is }\boxed{420}\:.[/tex]
[tex]\text{The factors of the ac product that add to 41 are }\:\boxed{20} \: \text{ and }\: \boxed{21}\:.[/tex]
[tex]35x^2+41x+12=\left( \: \boxed{5}\:x+\:\boxed{3}\:\right)\left(\: \boxed{7}\:x+\boxed{4}\:\right)[/tex]
Step-by-step explanation:
Given quadratic:
[tex]35x^2+41x+12[/tex]
Factoring quadratics by grouping
To factor a quadratic in the form [tex]ax^2+bx+c[/tex], find two numbers that multiply to ac and sum to b.
[tex]\implies ac=35 \cdot 12 = 420[/tex]
[tex]\implies b = 41[/tex]
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:
[tex]\implies 35x^2+20x+21x+12[/tex]
Factorize the first two terms and the last two terms separately:
[tex]\implies 5x(7x+4)+3(7x+4)[/tex]
Factor out the common term [tex](7x+4)[/tex] :
[tex]\implies (5x+3)(7x+4)[/tex]
Conclusion
[tex]\text{The ac product of }\: 35x^2+41x+12 \text{ is }\boxed{420}\:.[/tex]
[tex]\text{The factors of the ac product that add to 41 are }\:\boxed{20} \: \text{ and }\: \boxed{21}\:.[/tex]
[tex]35x^2+41x+12=\left( \: \boxed{5}\:x+\:\boxed{3}\:\right)\left(\: \boxed{7}\:x+\boxed{4}\:\right)[/tex]
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