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
@@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.
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.
That's not a quadratic tho, it's a cubic polynomial
It was quadratic after she factored out the x.
crystal clear!!! Thanks for your explanation...and excellent penmanship!
Nice method and clear video! Subscribed!
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
❌(❌-5)(2❌+3)
Thank you
Is there an explanation of the method using an a b c trinomial?
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.
@@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.
U made it look easy
Great Method! I’m having trouble solving 2x^2 - 6x - 10 with this method. What 2 factors of -20 sum to -6? 🤷♂️
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.
2x^2-6x-10=2(x^2-3x-5) :)