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Video 1 : Factoring Polynomials
One way to find factors of polynomials is to formed the algebraic long division. For example lets see x minus 3 is a factor of x cube minus seven x minus six? When dividing x minus 3 into x cube minus seven x minus six. First step the problem make a long division problem. There is you dividing x – 3 into x cube plus zero x square minus seven x minus six. Zero because there is no second degree term. Now you must ask yourself what times x give you x cube? Of, course x square, so you multiply x minus 3. By x square, which give you x cube minus 3 x square to get 3 x square. Bring it down next term negative seven x. Dividing x minus 3 into 3 x square minus seven x. Just looking at the first time 3 x square dividing x is 3x. Multiply x minus 3 by 3x. We can get 3 x square minus nine x. Subtracting you have 2x minus 6. Dividing 2 x minus 6 by x-3 which equals 2 and without a remainder. So the solution for a long division problem is x square plus 3 x plus 2. Since x minus 3 divide in x cube minus seven x minus 6. We now know x cube minus 7 x minus 6 equals ( x minus 3) times ( x square plus 3 x plus 2 ). The quadratic expression x square plus 3 plus 2 can be factored into ( x plus 1) times ( x plus 2). So, x cube minus 7 x minus 6 equals ( x minus 3 ) times ( x plus 1) times ( x plus 2 ). Substitution x cube minus 7x minus 6 to zero we get 0=(x-3)(x+1)(x+2) Thus either x-3=0 or x+1=0 or x +2=0 Solving of x we get x=3, x=-1, x=-2. The roots of x cube minus 7 x minus 6 are 3, -1, -2.
Conclusion:
*) 3 roots for this 3rd degree equation
*) 2nd degree equation always have at most 2 roots
*) 4th degree equation would have 4 or fewer roots, and so on.
*) The degree of polynomials equation always limits the number of roots.
Long division process for 3rd order Polynomial:
1. Find a partial quotient of x square, by dividing x into x cube to get x square
2. Multiply x square by the divisor and subtract the product from the dividend.
3. Repeat the process until you either “ clear it out “ or reach a remainder.

Video 2 : Solve The Problems
The next problem is question 13. The figure shows of graph of y equals g of x. If the function h is defined by h(x) = g(2x)+2, what is the value of h(1)? And we are looking for h of 1.
The information of the graph h(x)=g(2x)+2 now we are looking for h(1). We substitute h(1) into this equation. H(1)=g(2)+2 now we get g(2), see when x equals 2, y equals 1. So g(2) is 1. H(1)=g(2)+2, h(1)=1+2, so h(1)=3.

Next is the function with no function problem. This question 13 page 534. Let the function f be defined by F(x)=x+1, if 2f(p)=20, what is the value of f(3p)? We are looking f(3p) what is f when x = 3p?
First information is F(x)=x+1, 2f(p)=20 The figure of 3p start with this equation down here:
2f(p)=20, the, if we are divide by 2 so : f(p)=10. Then f(p) is just what is function f(x) of f(p) f(p)=p+1=10, then p=9. Is this right answer? No, is not. We looking for x=3p, x=27. We have an equation f(x)=x+1, f(27)=27+8=28. The last answer is 28.

Question 17.
In the x y – coordinate plane, the graph of x equals y square minus 4 intersect line l at ( 0,p ) and ( 5,t ). What is the greatest possible value of the slope of l? we’ll be looking for greatest m. The graph intersect in x = 4 line l intersect at ( 0,p ) and ( 5,t ). X is zero, y is p and when x = 5, y=t what is the possible slope for line l. What we are doing now for get the slope? m
equals (y-y1) over (x2-x1) The slope is going to be m=(t-p) over 5. Numerator is t-p we have x=y square minus 4. We can play again the point of intersect ( 0,p ) and ( 5,t ) to the equation x= x square minus 4.

Video 3 : Pre Calculus
Graph of a rational function which can have discontinuities because has polynomial in the denominator.
Is possible value x divide by 0
Example : f(x) equals (x plus 2) over (x minus 1) F(1)= The value become 1+2 over 1-1 equals 3 over zero. That is bad idea.
Graph f(1)=1 plus 2 over zero : Break in finction graph.
F(x) = ( x+ 2) over ( x-1) : insert 0
F(0)=0+2 over 0-1 equals -2. Insert 1 F(1)=1+2 over 1-1 equals 3 over 0 is impossible.
Rational functions don’t always work in this way! Take graph f(x) = 1 over ( x square plus 1 ). Not all rational functions will give zero in denominator because of the +1 ( never zero ).
Rational functions denominator can be zero.
Polynomial have smooth and unbroken curve and for rational function x : zero in the denominator. That impossible situation. A break can show up in two ways. A simply type break is missing point on the graph. Y = ( x square minus x minus 6 ) over ( x minus 3). The graph loose like this if x = 3 ( 3 square minus 3 minus 6 ) over (3 minus 3) equals 0 over 0. That is not possible, not feasible, and not allowed. So that is no way if x = 3. This is a typical example to the missing point syndrome. Y = ((3 square minus 3 minus 6 ) over (3 minus 3) equals 0 over 0. When you see result of 0 over 0 and also tell you direction by possible factor top and bottom of rational function and simplify. For example. Y = ( x square minus x minus 6 ) over ( x minus 3), Equals ( x minus 3 ) times ( x plus 2) over ( x minus 3) so, y = x+2.
video4.
Now we talking about the inverse function.
we can write this function becomes relation of f(x,y)=o, then function y=f(x). is a straight line with y intersect (-1) and x intersect (1/2). Look at the line y=x. That line intersect the graph of y=2x-1. So, we get
x=2x-1
1+x=2x
then 1=x
So, the intersection of the line y=x and y=2x-1 with x=1 is (1,1).