Factoring quadratics using the AC method

[@TateJenson]

crystal clear!!! Thanks for your explanation...and excellent penmanship!

[@buzzz241]

Nice method and clear video! Subscribed!

[@MrFrmartin]

How I factored this was: 1) x (2x2 - 7x - 15) 2) x (x^2 - 7x - 30) # multiply A & C 3) x (x - 10)(x+3) --> x (x - 10/2)(x+3/2) #Factor, then divide the constants by 'A' Answer: x = 0, 5, -3/2

[@fredrickeffi2143]

U made it look easy

[@phaganators]

Thank you

[@fr0sty_saiyan491]

That's not a quadratic tho, it's a cubic polynomial

[@davidwebster9788]

It was quadratic after she factored out the x.

[@jaggisaram4914]

❌(❌-5)(2❌+3)

[@buzzz241]

Great Method! I’m having trouble solving 2x^2 - 6x - 10 with this method. What 2 factors of -20 sum to -6? 🤷‍♂️

[@MathlessonswithLinda]

Keep in mind that not everything is factorable with integer solutions. You will need to use a different method, either completing the square or the quadratic formula.

[@user-xq8dc9ex5j]

2x^2-6x-10=2(x^2-3x-5) :)

[@davidwebster9788]

Is there an explanation of the method using an a b c trinomial?

[@MathlessonswithLinda]

Yes, it is this video and others like it. A trinomial is three terms, you only need to factor out the x as a GCF first.

[@davidwebster9788]

@@MathlessonswithLinda I was refering to quadratics and understand what you mean by factoring. What I mean is given this trinomial, A*x^2+B*x+C, is there a general rational for the method and choices you described. Or, why not just use the Quadratic Formula.