6.7 Pg.366 This ppt includes 7 slides consisting of a Review and 3 examples

 Find all the zeros:  f(x)=x 3 +x 2 -2x-2  Answer: -,,-1  F(x)= x 3 – 6x 2 – 15 x = (x + 4)(x – 5)(x – 5)  the zeros are: -4, 5, 5  5 is a repeated solution A polynomial to the nth degree will have n zeros.

 (should be 3 total! degree 3)  CT = ±1 ±2 ±3 ±4 ±6 ±8 ±12 ±16 ±24 ±48  LC±1  Graph the equation and you’ll see only 1 real zero:  Look in the table and you will find -3 is the only zero in the table, SO use synthetic division with -3   x = 0 x 2 = -16 x = ±√-16 = ±4i The three zeros are -3, 4i, -4i

 F(x)= (x-1)(x-(-2+i))(x-(-2-i))  F(x)= (x-1)(x+2-i)(x+2+i)  f(x)= (x-1){(x+2)-i} {(x+2)+i}  F(x)= (x-1){(x+2) 2 -i 2 } Foil  F(x)=(x-1)(x 2 + 4x + 4 –(-1))Take care of i 2  F(x)= (x-1)(x 2 + 4x )  F(x)= (x-1)(x 2 + 4x + 5)Multiply  F(x)= x 3 + 4x 2 + 5x – x 2 – 4x – 5  f(x)= x 3 + 3x 2 + x - 5

 Note: 2+i means 2-i is also a zero  F(x)= (x-4)(x-4)(x-(2+i))(x-(2-i))  F(x)= (x-4)(x-4)(x-2-i)(x-2+i)  F(x)= (x 2 – 8x +16)((x-2)-i)((x-2)+i)  F(x)= (x 2 – 8x +16)((x-2) 2 -i 2 )  F(x)= (x 2 – 8x +16)(x 2 – 4x + 4 –(-1))  F(x)= (x 2 – 8x +16)(x 2 - 4x + 5)  F(x)= x 4 –4x 3 +5x 2 –8x 3 +32x 2 -40x+16x 2 -64x+80  F(x)= x 4 -12x 3 +53x x+80

 Under y= type in the equation.  Go to second; calc; 2:zero  Left bound: you need to place the cursor to the left of the intersection and press enter.  Right bound: you need to place the cursor to the right of the intersection and press enter; and enter again.  At the bottom of the window “zero” will appear x = # This is your real zero.

 Assignments will be made in class and placed on the web page under lesson plans.