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Creating Polynomials Given the Zeros. What do we already know about polynomial functions? They are either ODD functions Linear y = 4x - 5 They are either EVEN functions Cubic y = 4x3 - 5 Fifth Power y = 4x5 –x + 5 Quadratics Quadratics y = 4x2 -2 5 y = 4x - 5 Quartics y = 4x4 - 5 We know that factoring and then solving those factors set equal to zero allows us to find possible x intercepts. TOOLS WE’VE USED GCF (x + )(x + ) The “6” step Grouping Cubic** p/q Long Division (works on all Factoring factors of any degree) Quadratic Formula Synthetic Division (works only with factors of degree 1) We know that solutions of polynomial functions can be rational, irrational or imaginary. X intercepts are real. Zeros are x-intercepts if they are real Zeros are solutions that let the polynomial equal 0 We have seen that imaginaries and square roots come in pairs ( + or -). If we factor a polynomial and end up with : (4x 8)( x 1) 0 2 2 We would get solutions of : 2 and i So we could CREATE a polynomial if we were given the polynomial’s zeros. Write a polynomial function f of least degree that has rational coefficients, a leading coefficient of 1 and the given zeros. -1, 2, 4 Step 1: Turn the zeros into factors. (x+1)(x- 2)(x- 4) Step 2: Multiply the factors together. 3 3 2 2 f(x) x =x - 5x- 5x +2x+2x + 8+ Step 3: Name it! 8 Write a polynomial function f of least degree that has rational coefficients, a leading coefficient of 1 and the given zeros. 2i, 3 Must remember thatthe “i”szeros and roots in pairs. Step 1: Turn intocome factors. 2i,2i, 3 , 3 ( x 2i )( x 2i )( x - 3 )( x 3 ) Step 2: Multiply factors. 3) f ( x)( x x 4)(xx -12 2 4 22 Write a polynomial function f of least degree that has rational coefficients, a leading coefficient of 1 and the given zeros. 2 i, 2 - 3 Step 1: Turn intocome factors. Must remember thatthe “i”s zeros and roots in pairs. 2 i, - 2 i, 2 - 3 ,2 3 ( x 2 i )( x 2 i )( x - 2 3 )( x 2 3 ) Step 2: Multiply factors. ( x 2 i )( x 2 i )( x - 2 3 )( x 2 3 ) x x 2 2 x i x x 2 2x ix 2x 4 2i -i x x -ix -2i -i12 2 (x + 4x + 5) x -2 3 x x x x2 -2x x -2 -2x 4 x x 3 x 32 3 2 (x - 3 2 3 -3 4x + 1) x2 2 x+ 4x + 5 4 x 4x3 5x2 3 2 -20x 4x -16x -4x 2 x 4x 5 1 f(x) = 4 2 x -10x -16x + 5 -3