## gjhfdfg 2 years ago Solve the polynomial inequality and graph the solution set on a number line. Express the solution set in interval notation.

1. gjhfdfg

x^2 - 4x - 12 ≤ 0

2. ZeHanz

First, solve x²-4x-12=0, by factoring as (x ....)(x ....)=0. Once you've found the zeroes, put these on the number line. They will divide the number line into three pieces. Because the graph of x²-4x-12 is a parabola with a minimum, the part of the number line between the two zeroes gives a negative value, the other parts a positive value. Mark the areas with "+ + +", "- - -" and "+ + +". Now you can see the solution of the inequality right in front of you: it is the middle part.

3. gjhfdfg

Im lost

4. ZeHanz

Don't be! Can you factor x²-4x-12=0?

5. gjhfdfg

x*x - 2*2*x-2*2*3 Like that?

6. ZeHanz

No, like (x + ...)(x + ...)=0. On the dots you need numbers a and b. Now a+b=-4 (that's the coefficient of the middle term -4x), and a*b = -12, the constant third term. Can you find a and b?

7. gjhfdfg

2 & 12 being a & b?

8. ZeHanz

No, because 2+12 = 14 instead of -4 and 2*12 = 24 instead of -12. If you take a = 2 and b = -6, you get 2+(-6)=-4 and 2*(-6)=-12, so: x²-4x-12=0 is the same as (x+2)(x-6)=0. Are you familiar with this process?

9. gjhfdfg

Kinda,

10. ZeHanz

OK, you probably need to practise this a few times, to get the hang of it. Now look at (x+2)(x-6)=0. It is the product of two numbers, x+2 and x-6. This product is 0. If you multiply two numbers and the outcome is 0, what does that tell you about these numbers?

11. gjhfdfg

That the product is 0 still? I dont know..

12. ZeHanz

If you have 4*6, 2*59, 63*456, you may not always be able to give the answer immediately, but one thing you do know: the answer is certainly NOT zero. Agreed?

13. gjhfdfg

Agreed,

14. ZeHanz

So suppose I have multiplied two numbers and the outcome IS zero, what do you know about (at least one of) these numbers?

15. gjhfdfg

I dont get what your asking for, it would just be zero

16. ZeHanz

It means that at least one of the numbers must be zero! 0*456=0, -34*0=0, 0*0=0, but 34*53 is not 0. This means, if you have (x+2)(x-6)=0, you can split this equation into two very easy ones: x+2=0 or x-6=0. Can you see their solutions?

17. gjhfdfg

No

18. ZeHanz

Can you solve the equation x+2=0?

19. gjhfdfg

-2

20. ZeHanz

OK, and in the same way, the solution of x-6=0 is 6. We now have the solutions of (x+2)(x-6)=0. Solutions are -2 and 6. Put them on a number line, like this: ------------|------------|----------- -2 6

21. gjhfdfg

wouldnt it be -6? instead of just 6

22. ZeHanz

No, because we need to set x=6 in the equation x-6=0 to make it true. Next step: We can also put zeroes above -2 and 6 to indicate the value of (x+2)(x-6) there: 0 0 ------------|------------|----------- -2 6

23. ZeHanz

Keep in mind we are trying to solve x² - 4x - 12 ≤ 0. We have indicated on the number line this: x² - 4x - 12 = 0. That is not enough, we also need to know what values of x will make the thing < 0.

24. ZeHanz

To find out what the sign (pos or neg) x²-4x-12 will get, just try a number between -2 and 6, e.g. x=0. 0² - 4*0 -12 = -12 <0, so negative there. If you try in the other parts of the line you get this:

25. gjhfdfg

Wouldnt that make them infinitys?

26. ZeHanz

If you put in very large numbers for x, the value of x^2 - 4x - 12 will also get very large, but we need not to worry about that, because we have to answer another question: What values of x make x^2 - 4x - 12 smaller than zero? The answer is visible in the image: between -2 and 6! So if I color all solutions red, I get this:

27. ZeHanz

Only one thing left: we have to write the solution (drawn in red) in interval notation...

28. ZeHanz

How do you write all the numbers from -2 to 6 (including -2 and 6!) as an interval?

29. gjhfdfg

(-2,6)

30. ZeHanz

That would exclude -2 and 6...

31. ZeHanz

To include them, you write [-2, 6]. One final word of advice: from the answers you've given I get the impression that you are not yet ready to solve problems like this. I really think you should practise elementary algebra skills. Nevertheless, I hope I have helped you a little...