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Finding Rational Zeros 6.6 pg. 359!. The rational zero 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 factor of constant term a 0 q factor of leading coefficient a n. n. =. Example 1:.
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The rational zero theorem … • If f(x)=anx + +a1x+a0 has integer coefficients, then every rational zero of f has the following form: pfactor of constant term a0 q factor of leading coefficient an n =
Example 1: • Find rational zeros of f(x)=x3+2x2-11x-12 • List possible LC=1 CT=-12 X= ±1/1,± 2/1, ± 3/1, ± 4/1, ± 6/1, ±12/1 • Test: 1 2 -11 -12 1 2 -11 -12 X=1 1 3 -8 x=-1 -1 -1 12 1 3 -8 -20 1 1 -12 0 • Since -1 is a zero: (x+1)(x2+x-12)=f(x) Factor: (x+1)(x-3)(x+4)=0 x=-1 x=3 x=-4
Extra Example 1: • Find rational zeros of: f(x)=x3-4x2-11x+30 • LC=1 CT=30 x= ±1/1, ± 2/1, ±3/1, ±5/1, ±6/1, ±10/1, ±15/1, ±30/1 • Test: 1 -4 -11 30 1 -4 -11 30 x=1 1 -3 -14 x=-1 -1 5 6 1 -3 -14 16 1 -5 -6 36 X=2 1 -4 -11 30 (x-2)(x2-2x-15)=0 2 -4 -30 (x-2)(x+3)(x-5)=0 1 -2 -15 0 x=2 x=-3 x=5
Example 2: • f(x)=10x4-3x3-29x2+5x+12 • List: LC=10 CT=12 x= ± 1/1, ± 2/1, ± 3/1, ± 4/1, ± 6/1, ±12/1, ± 3/2, ± 1/5, ± 2/5, ± 3/5, ± 6/5, ± 12/5, ± 1/10, ± 3/10, ± 12/10 • w/ so many –sketch graph on calculator and find reasonable solutions: x= -3/2, -3/5, 4/5, 3/2 Check: 10 -3 -29 5 12 x= -3/2 -15 27 3 -12 10 -18 -2 8 0 Yes it works * (x+3/2)(10x3-18x2-2x+8)* (x+3/2)(2)(5x3-9x2-x+4) -factor out GCF (2x+3)(5x3-9x2-x+4) -multiply 1st factor by 2 ____ __
Repeat finding zeros for: • g(x)=5x3-9x2-x+4 • LC=5 CT=4 x:±1, ±2, ±4, ±1/5, ±2/5, ±4/5 *The graph of original shows 4/5 may be: 5 -9 -1 4 x=4/5 4 -4 -4 5 -5 -5 0 (2x+3)(x-4/5)(5x2-5x-5)= (2x+3)(x-4/5)(5)(x2-x-1)= mult.2nd factor by 5 (2x+3)(5x-4)(x2-x-1)= -now use quadratic formula for the last factor- *-3/2, 4/5, 1± , 2 ____ __