SOLUTION: I ask the wrong question, sorry
factoring a quadratic with leading cofficient greater than one
2z^2-z-6
Thank you for the help, I would never get these with out this help
Algebra.Com
Question 91465: I ask the wrong question, sorry
factoring a quadratic with leading cofficient greater than one
2z^2-z-6
Thank you for the help, I would never get these with out this help
Found 2 solutions by stanbon, Earlsdon:
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
factoring a quadratic with leading cofficient greater than one
2z^2-z-6
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ac = 2*-6 = -12
b = -1
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Think of two numbers whose product is -12 and whose sum is -1
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The numbers are -4 and +3
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Replace -z by -4z+3z as follows:
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2z^2-4z+3z-6
Factor the 1st two and the last two terms separately:
2z(z-2)+3(z-2)
Factor again to get the final answer:
(z-2)(2z+3)
==================
Cheers,
Stan H.
Answer by Earlsdon(6294) (Show Source): You can put this solution on YOUR website!
Factor:
First, find the factors of the leading coefficient (that's the 2).
There is only one possibility and that's 1*2
So you can start by writing (2x + m)(x + n)
Now m*n must be equal to -6 (the consatnt term in the quadratic equation) so the possibilities are:
m*n = -6 So you know that m and n will have opposite signs
-1*6 = -6 (m = -1 and n = 6)
1*(-6) = -6 (m = 1 and n = -6)
-2*3 = -6 (m = -2 and n = 3)
2*(-3) = -6 (m = 2 and n = -3)
You also know that 2(n)+m = -1 (This is the coefficient of the z-term).
So we try:
(2z+3)(z-2) Using FOIL to multiply, we get:
Combining like-terms, we get:
So the factors are:
(2z+3)(z-2)
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