MEMORIAL UNIVERSITY OF NEWFOUNDLAND

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1 MEMORIAL UNIVERSITY OF NEWFOUNDLAND DEPARTMENT OF MATHEMATICS AND STATISTICS Section 5. Math 090 Fall 009 SOLUTIONS. a) Using long division of polynomials, we have x + x x x + ) x 4 4x + x + 0x x 4 6x + 9x Hence Qx) = x + x and Rx) = x + 7. b) By long division of polynomials, we have x 7x + 0x x 4x + 6x x + 4x x + 6x 9 x + 7 x 7x + 9x + x + ) x 4 4x + x + 0x x 4 + x 7x + x + 0x 7x 7x Hence Qx) = x 7x + 9x + and Rx) =.. a) By synthetic division with r =, we have Qx) = x x + 8 and R = 9. b) By synthetic division with r =, we have Qx) = x + 4x x + and R = 7. 9x + 0x 9x + 9x x x

2 . a) The possible rational roots are ±, ±, ±5 and ±0. Note that P ) = 4 but P ) = 0 is a root of P x) and x + ) is a factor. By synthetic division, we have P x) = x + )x + 4x 7x 0). Now for Qx) = x + 4x 7x 0 the possible rational roots are, ±, ±5 and ±0. We have Q ) = 0 is again a root of P x) and x + ) is again a factor. By synthetic division, we have and now P x) = x + ) x + x 0) = x + ) x + 5)x ). Hence the roots of P x) are, 5 and. b) The possible rational roots of P x) are ±, ±, ±4, ±, ±, ±, ± 4 and ±. Note that 6 P ) = 5 and P ) = 9 but P ) = 0. Thus is a root of P x) and x ) is a factor. By synthetic division, we have P x) = x )6x + x ) = x )x + )x ). Hence the roots of P x) are, and. c) The possible rational roots of P x) are ±, ±, ±5, ±0, ± and ± 5. Note that P ) = 6, P ) = 0, and P ) = but P ) = 0. Thus is a root of P x) and x + ) is a factor. By synthetic division, P x) = x + )x 9x + 4x 5). The possible rational roots of Qx) = x 9x + 4x 5 noting that the coefficients alternate in sign) are 5, and 5. We have Q5) = 90 but Q ) = 0 is a root of P x) and x ) is a factor. Again using synthetic division we get

3 P x) = x + ) x ) x 8x + 0) = x + ) x ) x 4x + 5). By the quadratic formula, if x 4x + 5 = 0 then x = 4 ± 6 0 = 4 ± 4 = 4 ± i = ± i. Hence the roots of P x) are,, + i and i. d) The possible rational roots of P x) are ±, ±, ±4, ±8 and ±. Note that P ) =, P ) = 9, but P ) = 0. Thus is a root of P x) and x ) is a factor. By synthetic division, P x) = x )x 4 x 6x + 6x + 4). The possible rational roots of Qx) = x 4 x 6x + 6x + 4 are ±, ±4 and ±. Note that P ) = 0 is again a root of P x) and x ) is one more a factor. By synthetic division, P x) = x ) x + x 4x ) = x ) [x x + ) x + )] = x ) x + )x ) = x ) x + )x )x + ), Hence the roots of P x) are,, and. e) The possible rational roots of P x) are,,,, and noting that the coefficients 9 9 alternate in sign and no negative roots can occur). Note that P ) = and P ) = 4, but P ) = 0. Thus is a root of P x) are x ) is a factor. By synthetic division,

4 4 In fact, note that P x) = = = x ) 9x x + 8x 6) x ) [x x ) + 6x )] x ) x )x + 6) = x ) x )x + ) = x ) x )x i)x + i). x ) = x this further simplifies to P x) = x ) x i)x + i). Either way, we see that the roots of P x) are, i and i. f) The possible rational roots of P x) are ±, ±, ±4 and ±8. Note that P ) = 0 and P ) = 6 but P ) = 0 is a root of P x) and x ) is a factor. By synthetic division, P x) = x )x 7x 4). The possible rational roots of Qx) = x 7x 4 are ± and ±4. Note that Q) = 0, Q ) = and Q4) = 8, but Q 4) = 0 4 is a root of P x) and x + 4) is a factor. By synthetic division, P x) = x )x + 4)x 4x ). If x 4x = 0 then, by the quadratic formula, x = 4 ± = 4 ± 0 = 4 ± 5 = ± 5. Hence the roots of P x) are, 4, + 5 and a) The possible rational roots of P x) are ±, ±, ±4, ±, ±, ± and ±. Note that P ) = 5 and P ) = 5, but P ) = 0 is a root of P x) and x ) is a factor.

5 5 By synthetic division, P x) = x )6x x x + ) = x )[6x x ) x )] = x )x )6x ) = x ) 4x )4x + ). b) The possible rational roots of P x) are ±, ±, ±, ±4, ±6, ±, ± and ±. Note that P ) = and P ) = 8, but P ) = 0 is a root of P x) and x ) is a factor. Then, by synthetic division, P x) = x )x + 9x + 7x 6). The possible rational roots of Qx) = x + 9x + 7x 6 are ±, ±, ±6, ± and ±. Note that Q) = 60 but Q ) = 0 is a root of P x) and x + ) is a factor. By synthetic division, P x) = x )x + )x + 5x ) = x )x + )x )x + ). c) The possible rational roots of P x) are ±, ±, ±4, ±8, ±, ±, ± 4 and ± 8. Note that P ) = 0, P ) = 0, and P ) = 60, but P ) = 0 is a root of P x) and x + ) is a factor. By synthetic division, P x) = x + )x x + x 4) = x + )[x x ) + 4x )] = x + )x )x + 4) = x + )x )x i)x + i).

6 6 d) The possible rational roots of P x) are ±, ±, ± and ±. Note that P ) = 0 is 4 a root of P x) and x ) is a factor. By synthetic division, P x) = x )4x 4 + 4x 7x 8x ). The possible rational roots of Qx) = 4x 4 + 4x 7x 8x are ±, ±, ± and ±. Note that Q) = 9, Q ) =, 4 Q) = 50, Q ) = 8, and Q ) ) = 7, but Q = 0 is a root of P x) and x + ) is a factor. Then by synthetic division, P x) = x ) x + ) 4x + x 8x 4) = x ) x + ) [x x + ) 4x + )] = x ) x + ) x + )x ) = x ) x + ) x + )x )x + ) = x )x + ) x )x + ). 5. a) Solving the given equation is equivalent to finding the roots of the polynomial P x) = x 4 + x x 7x. The possible rational roots of P x) are ±, ±, ± and ±. Note that P ) = 8 but P ) = 0 is a root of P x) and x + ) is a factor. By synthetic division, P x) = x+)x +x 4x ). The possible rational roots of Qx) = x +x 4x are, ±, ± and ±. Note that Q ) = 0 is again a root of P x) and x + ) is once more a factor. Then synthetic division yields 4 0 P x) = x + ) x x ) = x + ) x )x + ) = x + ) x ). Thus the lutions of the equation are x = and x =.

7 7 b) Solving the given equation is equivalent to find the roots of the polynomial P x) = 8x 5 +x 4 +4x +x +6x+. The possible rational roots of P x) are,, and 4 observing that all of the coefficients of P x) are positive). Note that P ) = 8 but P ) = 0 is a root of P x) and x + ) is a factor. By synthetic division, P x) = x + ) 8x 4 + 8x + 0x + 8x + ). The possible rational roots of Qx) = 8x 4 + 8x + 0x + 8x + are, and. Note that Q 4 8 ) = 0 is at least a double root of P x) and x + ) is again a factor. Synthetic division gives P x) = x + ) 8x + 4x + 8x + 4) = 4 x + ) x + x + x + ) = 4 x + [x ) x + ) + x + )] = 4 x + x + )x ) + ) = 4 x + x + )x i)x + i) ) = x + ) x i)x + i). Hence the lutions of the equation are x =, x = i and x = i. c) Solving this equation is equivalent to finding the roots of the polynomial P x) = x +x 0. The possible rational roots of P x) are ±, ±, ±5 and ±0. Note that P ) = 8 and P ) =, but P ) = 0 is a root of P x) and x ) is a factor. By synthetic division, P x) = x )x + x + 5). If we set x + x + 5 = 0 then the quadratic formula gives x = ± 4 0 = ± 6 ± 4i = = ± i. Thus the lutions of the equation are x =, x = + i and x = i.

8 8 6. a) Such a polynomial is P x) = x ) x + ) = x 4 5x + 6x + 4x 8. b) Note that if + i is a root, then is i. Therefore such a polynomial is P x) = x )x + i))x i)) = x )x + x + ) = x x 4x 6. c) First let s just find any polynomial with 4 and as roots. Since is a root of the polynomial, too is +. Hence one such polynomial is P x) = x+4)x ))x + )) = x+4)x 6x+) = x x x+4. However, observe that P 0) = 4. Since we want P 0) = 8 = 4, the polynomial we re actually looking for is P x) = P x) = x + 4x + 46x 8.

1) The line has a slope of ) The line passes through (2, 11) and. 6) r(x) = x + 4. From memory match each equation with its graph.

1) The line has a slope of ) The line passes through (2, 11) and. 6) r(x) = x + 4. From memory match each equation with its graph. Review Test 2 Math 1314 Name Write an equation of the line satisfying the given conditions. Write the answer in standard form. 1) The line has a slope of - 2 7 and contains the point (3, 1). Use the point-slope