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Stitz-Zeager_College_Algebra_e-book

# Since 3 173 the two zeros match what expected from

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Unformatted text preview: ceed as before −2 1 00 8 ↓ −2 4 −8 1 −2 4 0 We get the quotient q (x) = x2 − 2x + 4 and the remainder r(x) = 0. Relating the dividend, quotient and remainder gives x3 + 8 = (x + 2) x2 − 2x + 4 . 3. To divide 4 − 8x − 12x2 by 2x − 3, two things must be done. First, we write the dividend in descending powers of x as −12x2 − 8x + 4. Second, since synthetic division works only 3 for factors of the form x − c, we factor 2x − 3 as 2 x − 2 . Our strategy is to ﬁrst divide 2 − 8x + 4 by 2, to get −6x2 − 4x + 2. Next, we divide by x − 3 . The tableau becomes −12x 2 3 2 −6 −4 2 ↓ −9 − 39 2 −6 −13 − 35 2 3 From this, we get −6x2 − 4x + 2 = x − 2 (−6x − 13) − 35 . Multiplying both sides by 2 and 2 distributing gives −12x2 − 8x + 4 = (2x − 3) (−6x − 13) − 35. At this stage, we have written −12x2 − 8x + 4 in the form (2x − 3)q (x) + r(x), but how can we be sure the quotient polynomial is −6x − 13 and the remainder is −35? The answer is the word ‘unique’ in Theorem 3.4. The theorem states that there is only one way to d...
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