Finding Rational Zeros

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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:

p factor of constant term a0

q factor of leading coefficient an

n …

=

Example 1:• Find rational zeros of f(x)=x3+2x2-11x-121. List possible LC=1 CT=-12 X= ±1/1,± 2/1, ± 3/1, ± 4/1, ± 6/1, ±12/12. Test: 1 2 -11 -12 1 2 -11 -12X=1 1 3 -8 x=-1 -1 -1 12 1 3 -8 -20 1 1 -12 03. 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+301. LC=1 CT=30x= ±1/1, ± 2/1, ±3/1, ±5/1, ±6/1, ±10/1, ±15/1, ±30/12. 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

1. 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

2. 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

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Repeat finding zeros for:• g(x)=5x3-9x2-x+41. 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 4x=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 quad for last- *-3/2, 4/5, 1± , 1- . 2 2

5

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5

Assignment