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
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.
@@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.
crystal clear!!! Thanks for your explanation...and excellent penmanship!
Nice method and clear video! Subscribed!
U made it look easy
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
Thank you
That's not a quadratic tho, it's a cubic polynomial
It was quadratic after she factored out the x.
❌(❌-5)(2❌+3)
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) :)
@@MathlessonswithLinda did u know I invented the quadratic formula?
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.