Adv Alegbra HW Solutions 146

# Adv Alegbra HW Solutions 146 - 146 5.13 Find the roots of x...

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146 5.13 Find the roots of x 3 6 x + 4. Solution. We have q =− 6, r = 4, D =− 16, and α 3 =− 2 + 2 i .UseDe Moivre s theorem to f nd cube roots of 2 + 2 i : ( 1 + i ) 3 =− 2 + 2 i . Hence, α = 1 + i , β = 1 i , and a root of the polynomial is α + β = 2. The other two roots can be found by two methods. We can use the cubic formula to obtain: ω( 1 + i ) + ω 2 ( 1 i ) and ω 2 ( 1 + i ) + ω( 1 i ), or one can use the division algorithm and the quadratic formula: x 3 6 x + 4 = ( x 2 )( x 2 + 2 x 2 ), so that the other roots are 1 ± 3. It is not clear from the cubic for- mula that the roots are all real; however, 27 r 2 + 4 q 3 =− 432 < 0, and Proposition 5.9 applies.) 5.14 Find the roots of x 4 15 x 2 20 x 6. Solution. This is a realistic problem; the implementation of the formula is rather long (but see the remark at the end of this solution). Set Q =− 15, R =− 20, and S =− 6. Then j 6 + 2 Qj 4 + ( Q 2 4 S ) j 2 R 2 = j 6 30 j 4 + 249 j 2 400 . We now make this cubic (in j 2 ) reduced with the substitution j 2 = y + 10, obtaining F ( y ) = y 3 51 y + 90 . Apply the cubic formula:
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