121616949-math.185

# 121616949-math.185 - and to factor the quadratic if it is...

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8.5 Rational Functions 171 EXAMPLE 8.15 Find Z t 3 (3 - 2 t ) 5 dx. Using the substitution u = 3 - 2 t we get Z t 3 (3 - 2 t ) 5 dx = 1 - 2 Z u - 3 - 2 · 3 u 5 du = 1 16 Z u 3 - 9 u 2 + 27 u - 27 u 5 du = 1 16 Z u - 2 - 9 u - 3 + 27 u - 4 - 27 u - 5 du = 1 16 u - 1 - 1 - 9 u - 2 - 2 + 27 u - 3 - 3 - 27 u - 4 - 4 = 1 16 (3 - 2 t ) - 1 - 1 - 9(3 - 2 t ) - 2 - 2 + 27(3 - 2 t ) - 3 - 3 - 27(3 - 2 t ) - 4 - 4 = - 1 16(3 - 2 t ) + 9 32(3 - 2 t ) 2 - 9 16(3 - 2 t ) 3 + 27 64(3 - 2 t ) 4 We now proceed to the case when the denominator is a quadratic polynomial. We can always factor out the coefficient of x 2 and put it outside the integral, so we can assume that the denominator has the form x 2 + bx + c . There are three possible cases, depending on how the quadratic factors: either x 2 + bx + c = ( x - r )( x - s ), x 2 + bx + c = ( x - r ) 2 , or it doesn’t factor. We can use the quadratic formula to decide which of these we have,
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Unformatted text preview: and to factor the quadratic if it is possible. EXAMPLE 8.16 Determine whether x 2 + x + 1 factors, and factor it if possible. The quadratic formula tells us that x 2 + x + 1 = 0 when x =-1 ± √ 1-4 2 . Since there is no square root of-3, this quadratic does not factor. EXAMPLE 8.17 Determine whether x 2-x-1 factors, and factor it if possible. The quadratic formula tells us that x 2-x-1 = 0 when x = 1 ± √ 1 + 4 2 = 1 ± √ 5 2 . Therefore x 2-x-1 = ˆ x-1 + √ 5 2 !ˆ x-1-√ 5 2 ! ....
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