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Find the real zeros of the function

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f(x)=6x^3-49x^2+20x-2?  

asked Sep 5, 2014 in ALGEBRA 1 by anonymous

2 Answers

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The function f (x ) = 6x 3 - 49x 2 + 20x  - 2

Identify Rational Zeros  

Usually it is not practical to test all possible zeros of a polynomial function using only synthetic substitution. The Rational Zero Theorem can be used for finding the some possible zeros to test.

 6x 3 - 49x 2 + 20x  - 2 = 0

If p/ is a rational zero, then p  is a factor of 2 and is a factor of 6.

The possible values of p  are   ± 1 and  ± 2.

The possible values for q  are ± 1, ± 2, ± 3 and ± 6.

So, p/q  = ± 1,  ± 2, ± 1/2, ± 1/3, ± 2/3, ± 1/6.

Make a table for the synthetic division and test possible  zeros.

p/q 6 -49 20 -2
1 6 -43 -23 -25
-1 6 -55 75 -77
2 6 -37 -54 -110
-2 6 -61 142 -286
1/2 6 -45 -5/2 -7
-1/2 6 -52 46 -25
1/3 6 -47 13/3 -5/9
1/6 6 -48 12 0

 

answered Sep 5, 2014 by david Expert
0 votes

Contd...

Since f (1/6) = 0, x = 1/6 is a zero. The depressed polynomial is   6x2 - 48x + 12 = 0

x2 - 8x + 2 = 0

Since the depressed polynomial of this zero, x2 - 8x + 2 , is quadratic, use the Quadratic Formula to find the roots of the related quadratic equation

x2 - 8x + 2  = 0

Compare it to image

Roots are

image

image

image

image

image

image

Real zeros are 1/6 , 7.74 and 0.26.

answered Sep 5, 2014 by david Expert

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