Question

solve by factoring
6t² + t – 5 = 0

Answer 1

Multiply the leading coefficient to the constant term. \[\color{red}6t^2+t\color{red}{-5}=0\]\[t^2+t-30=0\] find 2 numbers that multiply to give you 30 but add to give you 1.

Answer 2

What would be some factors of 30?

Answer 3

-6and 5

Answer 4

No, \((t-6)(t+5) \ne t^2+t+30\)

Answer 5

-5 and 6

Answer 6

Ok.

Answer 7

Therefore, \((t+6)(t-5)=0\)

Now since you used your leading coefficient to reduce your quadratic into something solvable, divide each of your factors by that term and reduce whichever fraction you cannot. \[\left(t+\color{red}{\frac{6}{6}}\right)\left(t-\frac{5}{6}\right)\] You can reduce the one that is highlighted in red but you cannot reduce the other. Simplify the second factor by multiplying the 6 in the denominator across to the variable.

Answer 8

\[\huge \boxed{(t+1)(6t-5)}\]