what type of polynomial is this
Answers
Answer:
2x^2-3x-9=0 can be factored into (2x+3) (x-3) =0, with x being equal to -(3/2) or 3. How do I get to that answer, and why isn't it just x=-3 or +3 as it usually is with quadratics?
For a quadratic of the form ax2+bx+c , where a is not 1, I like to start factoring by using the ac method.
For the ac method, you multiply the 2 numbers represented by a and c . In this specific case, we have 2x2−3x−9 , therefore we have a=2 , b=−3 , and c=−9 . Therefore, for this problem ac=(2)(−9)=−18 .
After multiplying a and c together, you need to consider if there are 2 numbers that multiply to that same value, but they add to make the number b . Therefore, we need to find 2 numbers that multiply to -18, but add to -3.
The first thing I notice, is they need to multiply to make a negative number, so we have one number that is negative and one number that is positive. The second thing I notice is -3 is a negative number also, so the negative number will have a higher absolute value than the positive number, in other words the negative number is “larger” than the positive number.
Then I consider the numbers that can multiply to 18. In this case, the choices are 1 and 18, 2 and 9, and 3 and 6. Again recall that the larger number needs to be negative, so that means we have 1 and -18, 2 and -9, or 3 and -6. Adding those 2 pairs together, the only one that gives us -3 is the pair 3 and -6.
Now that we have the 2 numbers that multiply to ac and add to b , we use those 2 numbers to split up the b into 2 separate terms.
2x2−3x−9
2x2+3x−6x−9
Now we can use factoring by grouping. This means we put parentheses around the first 2 terms, and we put parentheses around the final 2 terms. Any subtraction can be rewritten as adding a negative.
(2x2+3x)+(−6x−9)
Now we factor each of those groups individually. If the first term is negative, try to factor out a negative number. In the first set of parentheses, 2 and 3 do not share a common factor, but x2 and x do. Therefore, we can factor out an x there.
For the second set of parentheses, the first term is negative, so I will factor out a negative. In addition, both 6 and 9 can be divided by 3, so I can factor out a 3. That gives the following factoring:
x(2x+3)−3(2x+3)
Now we have 2 terms and we need to see if we can factor again. Notice that both terms have a (2x+3) in them, so we can factor that whole binomial out. That leaves us with an x for the first term, and a −3 for the second term:
(2x+3)(x−3)
That gives us the factoring we were looking for.
Recall that this was an equation, so we actually have (2x+3)(x−3)=0 . From here, we can use the zero-product property. If you are multiplying 2 things together and get a 0, either the first thing must be 0, or the second thing must be 0. Therefore we get 2 different equations from this:
2x+3=0 or x−3=0
Now we solve each of those separately.
2x=−3 or x=3
x=−32 or x=3 .
Therefore, that is our solution: x∈{−32,3}
As to why we do not have 2 identical numbers where one is positive and one is negative, the only way that would happen is if you had a difference of squares. For example, if you have the equation x2−a2=0 , then the 2 solutions would be x=a or x=−a .
Step-by-step explanation:
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