Andrew M. answered • 06/28/15
Tutor
New to Wyzant
Mathematics - Algebra a Specialty / F.I.T. Grad - B.S. w/Honors
a.
(x2-4x-12)(2x+3)2/(2x2-9x-18)
Factor the numerator: (x2-4x-12)
factors of -12 that add to -4 are -6, 2 ... so this factors to (x-6)(x+2)
Factor the denominator: 2x2-9x-18 factor by grouping...
multiply the coefficient of square term (2) by the constant (-18)... 2(-18) = -36
Factors of -36 that add to -9 are -12, 3 so break up -9x to -12x + 3x giving: 2x2-12x+3x-18)
= [(x-6)(x+2)(2x+3)(2x+3)]/(2x2-12x+3x-18)
= [(x-6)(x+2)(2x+3)(2x+3)]/[(2x(x-6) + 3(x-6)]
= [(x-6)(x+2)(2x+3)(2x+3)]/[(2x+3)(x-6)]
We can now cancel like terms from numerator and denominator
= (x+2)(2x+3)
b. Our final answer expanded back out; (x+2)(2x+3) = 2x2+7x+6
Let's see why this must be positive. In order to be negative we have:
2x2+7x+6<0
(x+2)(2x+3)<0
Either x+2>0 and 2x+3<0 or x+2<0 and 2x+3>0
x>-2 and x<-3/2 x<-2 and x>-3/2
In neither case is there any answer that fits both criteria so the answer to
2x2+7x+6<0 is the empty set
Now... As for why the answer cannot be -3/2 or 6 we have to look at the denominator of
the original problem. Remember we cannot divide by zero so we set the denominator
equal to zero to see what answers would make the equation undefined and thus create holes
in the graph
2x2-9x-18 = (2x+3)(x-6)=0
2x-3 = 0 x-6 = 0
x = 3/2 x=6
Either x= 3/2 or x=6 would put a zero in the denominator of the original function
Thus x cannot equal either 3/2 or 6
Marietta H.
I understand the positive bit, but not sure about the rest of b). Why did you not minus the 7x from the quadratic equation before factorising it to fine solutions for x? And in the 2x-3=0, x = 3/2 not -3/2?
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06/29/15
Andrew M.
For part b.. We need to show that 2x2 + 7x + 6=0 must be positive. To do that it is not enough to show that there are positive answers that work, we must also show that there are no negative answers that work. Originally I showed that there were no negative answers that worked because we cannot have x<-2 and x>-3/2. No numbers fit the criteria so there are no negative answers. In my comment I went on to show that there are positive numbers that work.
About the x = -3/2 instead of 3/2
It works out like this:
We have 2x+3>0 to move the 3 over we must subtract if from both sides
2x+3-3>0-3
2x>-3
We now have to isolate the x term by dividing out the 2. We only switch the inequality sign if we multiply or divide both sides by a negative number.... 2 is positive so the inequality remains as >
2x/2 > -3/2
x>-3/2
***************************
Factoring the quadratic 2x2+ 7x +6
It would do me no good to subtract 7x from both sides. I would get
2x2+6 = -7x
2(x2+3) = -7x
x2+ 3 = -7x/2
Now what? .... This does not work.
When looking at a quadratic of ax2 + bx + c which we wish to factor, and a≠1, we try to factor by grouping.
1) look at the coefficient 'a" of the squared term... in this case a= 2 ... and the constant "c" .. in this case 6
Multiply them together .... 2(6) = 12
2) From the product of a(c) try to find factors that will add to the coefficient "b" of the x term... in this case 7
the factors of 12 that add to 7 are 4 and 3.
3) Using those factors split the x term into two terms ... 4x + 3x .... and reintroduce them into the quadratic..
2x2 + 4x + 3x + 6 or 2x2 + 3x + 4x + 6
Look at the two quadratics ... They are actually the same. But, we want to be able to factor two parentheses out of this so we must group them by those that have factors in common.... In this case 2 and 4 have a common factor of 2 but 2 and 3 don't... So I will use 2x2+4x+3x+6
Isolate the first two terms and the last two terms....(2x2+4x) + (3x+6)
I can factor both parts of that into
2x(x+2) + 3(x+2)
What I have here is a(b) + c(b) = (a+c)(b)
This means I can rewrite this as (2x+3)((x+2)
I sincerely hope this helps.
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06/29/15
Marietta H.
Okay, I get it now! Thanks for your help.
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06/29/15
Andrew M.
06/28/15