M1410205Part3

M1410205Part3 - THE QF METHOD FOR FACTORING QUADRATICS ()...

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THE “QF” METHOD FOR FACTORING QUADRATICS Remember our old friend f x ( ) = 4 x 3 5 x 2 7 x + 2 . Let’s say we want to factor this completely over C . From Part B on the Rational Zero Test, we found a list of candidates for rational zeros. It turned out that 2 was, in fact, a zero. Therefore, by the Factor Theorem, x 2 ( ) was a factor of f x ( ) . After performing Synthetic Division, we found that: 4 x 3 5 x 2 7 x + 2 = x 2 ( ) 4 x 2 + 3 x 1 ( ) Trial-and-error can be used to factor the quadratic factor, but this method makes some people nervous. There is a more systematic alternative offered to us by the Quadratic Formula (QF). Ordinarily, we factor before finding zeros, but we will reverse that here. Use the QF to find the zeros of 4 x 2 + 3 x 1 ; in other words, solve 4 x 2 + 3 x 1 = 0 . Observe: a = 4 , b = 3 , and c = 1 . x = b ± b 2 4 ac 2 a = 3 ( ) ± 3 ( ) 2 4 4 ( ) 1 ( ) 2 4 ( ) = 3 ± 25 8 = 3 ± 5 8 x = 3 + 5 8 = 2 8 = 1 4 or x = 3 5 8 = 8 8 = 1 The zeros of 4 x 2 + 3 x 1 are 1 4 and 1 .
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Remember that the leading coefficient of 4 x 2 + 3 x 1 was a n = 4 . An LFT Form of 4 x 2 + 3 x 1 is, therefore: 4 x 2 + 3 x 1 = 4 x 1 4
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M1410205Part3 - THE QF METHOD FOR FACTORING QUADRATICS ()...

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