Solutions to the Worksheet on Polynomials and Rational Functions Math 141 1 Roots of Polynomials A Indicate the multiplicity of the roots of the polynomialh(x) = (x 1) ( x) 3( x +x+1 ) B Check the remainder theorem for the following division of the form p(x) x a, by evaluating p(a), and also by performing the division and finding the remainder: 3x 4 x 3 x +x 1 x+1 C Knowing that 3 is a root of the polynomial f (x) = x 3 7x +x+6, find the remaining roots Hint If a is the root of the cubic polynomial f(x), then f(x) = (x a)g(x), where g(x) is a quadratic polynomial. Solutions: A The polynomial is almost completely factored (we only need to observe that x +x+1 = (x+1), hence its roots are explicitly given: 1, with multiplicity, with multiplicity 3, and 1 with multiplicity. We can infer the behavior of the graph near each root from this information. The graph will look like 1
1 Roots of Polynomials B We are dividing3x 4 x 3 x +x 1 byx ( 1). Hence, the remainder should be equal to 3 ( 1) 4 ( 1) 3 ( 1) + ( 1) 1 = 3+1 1 1= 0. In fact, 3x 4 x 3 x +x 1 = (x+1) ( ax 3 +bx +cx+d ) (there should be no remainder, as we found out) means 3x 4 x 3 x +x 1 = ax 4 +(b+a)x 3 +(c+b)x +(d+c)x+d consistently. a = 3,b+a = b+3 = 1,b = 4,c+b = c 4 = 1,c = 3,d+c = d+3 =,d = 1 C Indeed, 3 is a root of x 3 7x +x+6. Hence, this polynomial is divided evenly (no remainder) by x 3. We find the quotient by division: x 3 7x +x+6 = (x 3) ( ax +bx+c ) x 3 7x +x+6 = ax 3 +(b 3a)x +(c 3b)x 3c a =,b 3a = b 6 = 7,b = 1,c 3b = c+3 = 1,c =, 3c = 6 Hence, our polynomial factors like x 3 7x +x+6 = (x 3) ( x x ) The remaining two roots of our polynomial are then Here is a graph: 1± 1+16 4 = 1 4 ± 17 4
Features 3 Features A Describe as many features as you can of the graphs of the following functions: 1. 5x 5 +3x 3 x x+1. x 6 +3x 3 x +x 1 3. a(x) = 3x+ 1 x x 4. b(x) = 3x 3 4x 4x 5. f(x) g(x), where f (x) = x3 +x 1,g(x) = x x 3 Hint: f(x) has an obvious root. Solutions For the first two, questions the solution is straightforward: for large x the highest power dominates, and for small x it is the lowest to set the tone. Hence, 1. The tail behavior is dictated by 5x 5, an odd power with a negative coefficient. Hence, for x towards the graph will be towards large positive values ( towards + ), and for x towards + it will go towards extreme negative values ( towards ). Near x = 0, the graph will look like 1 x, that is, its tangent at (0,1) will have slope 1. Similarly, the tail behavior is determined by x 6, an even power with a positive coefficient, so that the graph goes towards + in both directions. Neat x = 0, it will behave like x 1, crossing the y axis at (0, 1), with a positive slope of 1 Note: we could go farther about the behavior near x = 0, and note that in case 1, the next term is x, hence the graph (which is decreasing), is similar to that of a parabola open down. Hence, it will run to the left of the tangent.
Features 4 Y +10 +8 +6 +4 + 7.0 5.0 3.0 1.0 +.0 +4.0 +6.0 X 4 6 8 10 In case the term is again x. Since the graph is increasing here, it will lie to the right of the tangent line. Y +10 +8 +6 +4 + 7.0 5.0 3.0 1.0 +.0 +4.0 +6.0 X 4 6 8 10
Features 5 3. a(x) is obviously zero when x = 3, and its domain is the whole real line, without the two roots of its denominator: 1 and 1. This also tells us that x = 1 and x = 1 are vertical asymptotes. The end behavior is also clear, since the degree of the numerator is less than the degree of the denominator: y = 0 is a horizontal asymptote. To graph by hand the function, it is useful to look at how its sign changes, as we cross zeros and asymptotes. This is determined by the signs of the numerator and denominator. The numerator is negative for x < 3, and positive for x > 3. The denominator is negative for x < 1, and x > 1 (it is an open down quadratic function). Hence, we have that a is positive (negative/negative) for x < 3. Zero at x = 3 negative (positive/negative) for 3 x = 1 < x < 1. Vertical asymptote at positive (positive/positive) for 1 < x < 1. Vertical asymptote at x = 1 negative (positive/negative) for x > 1 This also shows that f approaches zero from above, as x tends towards, and from below when x tends towards +
Features 6 4. Similarly, b has a zero at x = 3, and vertical asymptotes at 3 and 1. The end behavior is similar to that of a. To get more information, we repeat the analysis we did on a, and observe that The numerator is negative for x < 3, and positive for x > 3, while the denominator is (again, an open down function) negative for x < 3, and x > 1. Hence, b is positive for x < 3. Vertical asymptote at x = 3 negative for 3 < x < 1. Vertical asymptote at x = 1 positive for 1 < x < 3. Zero at x = 3 negative for x > 3 The approach to zero at the ends is similar to that of f 5. With a little observation, we can see that f ( 1) = 0. Dividing f by x+1 we find its remaining roots, by applying the quadratic formula to the quotient: x 3 +x 1 = (x+1) ( ax +bx+c ) x 3 +x 1 = ax 3 +(b+a)x +(c+b)x+c a = 1,b+a = b+1 =,b = 1,c+b = c+1 = 0,c = 1 x 3 +x 1 = (x+1) ( x +x 1 ) and the two remaining roots are 1± 1+4 = 1 ± 5
Features 7 These three roots are, potentially, also the zeros of f g, obviously. However, any of them could not belong to the domain. The domain excludes the roots of g: ± 4+1 = 1± that are at x = 1 and x = 3. We see that 1 is not a zero of our function, since it does not belong to the domain. x = 3 will be an asymptote, but x = 1 will not, since, as x is very close to 1, the denominator is very small, but so is the numerator. In fact, if we factor x+1 from both, we have (x+1) ( x +x 1 ) (x+1)(x 3) so that if x 1, this will look very much like 1 (x+1) 4 (x+1) (compare the discussion in the book about the behavior of a polynomial near a root). As x gets closer and closer to 1, the graph does not run away f g gets closer and closer to 1 4, without reaching it, since x = 1 is not in the domain. It is useful to repeat the discussion of the signs to get a handle on the graph. Thus we have: f(x) being cubic with a positive coefficient, and knowing its roots, we can say it is negative for x < 1+ 5, positive for 1+ 5 < x < 1, negative for 1 < x < 5 1, positive for x > 5 1 (if you are curious, 1+ 5 1.618, 5 1 0.618 g being quadratic, open up, will be positive for x < 1, and x > 3, negative between the two Combining the discussion, we have that f g will be negative for x < 1+ 5, positive for 1+ 5 < 1, and for 1 < x < 5 1 (since both numerator and denominator change sign at x = 1), negative for 5 1 < x < 3, and positive for x > 3 To determine the end behavior, it is useful to determine the quotient (the remainder is not useful in this respect). We can use long division or indeterminate coefficients. Indeterminate coefficients works like this: x 3 +x 1 = ( x x 3 ) (ax+b)+cx+d = ax 3 +bx ax bx 3ax 3b+cx+d= from which we get = ax 3 +(b a)x (b+3a c)x 3b+d a = 1,b a = b = b = 4,b+3a c= 8+3 c = 0 c = 11, 3b+d = 1+d = 1 d = 11
Features 8 and we find x 3 +x 1 x x 3 = x+4+ 11x 11 x x 3 Long division yields the same result, obviously: x +4 x x 3 x 3 +x 1 x 3 x 3x 4x 3x 1 4x 8x 1 11x +11 We see that x + 4 will be an oblique asymptote. This is consistent with the sign we found for f g at both ends. One last simple observation is that the y intercept is at y = 1 3. Summing up we can sketch a graph that will look like this: (of course the computer cannot point out the hole in the graph at ( 1, 1 4) ). The vertical line is the asymptote at x = 3.
Features 9 B Looking at the following graphs of polynomial functions, fill in the following data: The degree of the polynomial is even or odd The degree of the polynomial is at least... The degree of this polynomial has to be even (with a positive leading coefficient), and not less than 4, given the three turning points. This polynomial has an odd degree (with negative leading coefficient), no less than three. Note: As a matter of fact, the graphs are of a 4th and 3rd degree polynomial respectively. However, given the imprecision of a picture, and the limited range of values pictured, it would be impossible to rule out higher degrees.
3 Operations 10 3 Operations Referring to the functions a,b,f,g defined in point, write (in nice order: all polynomials should be written in order of descending exponent) the result of 1. f (x) g(x). f (x) g(x) 3. a(x) b(x) Solutions: These are straightforward algebra exercises: 1. Subtracting requires care only in keeping track of the signs: x 3 x 1 ( x x 3 ) = x 3 x x +x 1+3= x 3 3x +x+. Multiplying through, thanks to the distributive rule, ( x 3 +x 1 )( x x 3 ) = x 5 x 4 3x 3 +x 4 4x 3 6x x +x+3 = = x 5 7x 3 7x +x+3 3. It is best to use an LCD, since this keeps the degrees of the polynomials as low as possible.. Since 1 x x = (1 x)(x+1), and 3 4x 4x = (1 x)(x+3), the LCD is (1 x)(x+1)(x+3) = 3 x 8x 4x 3. Consequently f(x) g(x) = (3x+)(x+3) (3x )(x+1) 4x 3 8x x+3 = 6x +13x+6 ( 3x +x ) 4x 3 8x = x+3 3x +1x+8 4x 3 8x x+3 =