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4.4 Finding Rational Zeros

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The rational root theorem If f(x)=a n x + +a 1 x+a 0 has integer coefficients, then every rational zero of f has the following form: p factors of constant term a 0 q factors of leading coefficient a n n … =

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Example 1: Find rational roots of f(x)=x 3 +2x 2 -11x-12 1.List possible q=1(1) p=-12 (1,2,3,4,6,12) X= ±1/1,± 2/1, ± 3/1, ± 4/1, ± 6/1, ±12/1 2.Test: X= x= Since -1 is a zero: (x+1)(x 2 +x-12)=f(x) Factor: (x+1)(x-3)(x+4)=0 x=-1 x=3 x=-4 You can use a graphing calculator to narrow your choices before you test!

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Extra Example 1: Find rational zeros of: f(x)=x 3 -4x 2 -11x q=1 (1) p=30 (1,2,3,5,6,10,30) x= ±1/1, ± 2/1, ±3/1, ±5/1, ±6/1, ±10/1, ±15/1, ±30/1 2.Test: x= x= X= (x-2)(x 2 -2x-15)= (x-2)(x+3)(x-5)= x=2 x=-3 x=5

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Example 2: f(x)=10x 4 -3x 3 -29x 2 +5x+12 1.List: q=10 (1,2,5,10) p=12 (1,2,3,4,6,12) x= ± 1/1, ± 2/1, ± 3/1, ± 4/1, ± 6/1, ±12/1, ± ½, ± 3/2, ± 1/5, ± 2/5, ± 3/5, ± 4/5, ± 6/5, ± 12/5, ± 1/10, ± 3/10, ± 12/10 2.w/ so many – sketch graph on calculator and find reasonable solutions: x= -3/2, -3/5, 4/5, 3/2 Check: x= -3/ Yes it works * (x+3/2)(10x 3 -18x 2 -2x+8)* (x+3/2)(2)(5x 3 -9x 2 -x+4) -factor out GCF (2x+3)(5x 3 -9x 2 -x+4) -multiply 1 st factor by 2 __ ____

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Descartes’ Rule of Signs The number of positive real zeros = the number of sign changes in f(x) or less than by an even #. NO YES There are four changes. So there may be 4, 2 or 0 positive real zeros.

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Descartes’ Rule of Signs cont. The number of negative real zeros = the number of sign changes in f(-x) or less than by an even #. YES NO There is one change. So there is 1 negative real zero.

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Find the number of positive and negative real zeros. Then determine the rational zeros for g(x)= 5x 3 - 9x 2 – x + 4 There are 2 positive and 1 negative real zero 1. LC=5 CT=4 x:±1, ±2, ±4, ±1/5, ±2/5, ±4/5 *The graph of original shows 4/5 may be: x=4/ (x-4/5)(5x 2 -5x-5)= (x-4/5)(5)(x 2 -x-1)= mult.2 nd factor by 5 (5x-4)(x 2 -x-1)= -now use quad for last- * 4/5, 1+, __ ____

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