PreCalculus Chapter 2 Practice Answers

Transcription

1 PreCalculus Chapter 2 Practice Answers 1. To find the x and y intercepts, let x = 0 to find the y intercept and let y = 0 to find the x. y = x 2 5x + 4 y = (0) 2 5(0) + 4 y = 4, when x = 0 (0,4) 0 = x 2 5x = ( x 4)( x 1) x = 1, 4 when y = 0 (1,0),(4,0) 2. If possible, always remove the Greatest Common Factor (GCF) first. 0 = x 3 3x 2 4x Factor (always remove the 0 = x( x 2 3x 4) greatest common factor first) 0 = x( x 4)( x +1) x = 0, 1,4 Use Principal of Zero Products 3. Note this equation is quadratic-like and as such can be factored and solved like a quadratic. You may be able to factor it directly; if not, use substitution techniques. Let u = x 2 then u 2 = x 4 x 4 5x 2 36 = 0 u 2 5u 36 = 0 (u 9)(u + 4) = 0 u = 9 or u = 4 x 2 = 9 or x 2 = 4 x = ±3 Using substitution makes the problem a simple quadratic that is easy to factor. Solve for u, then re-substitute the x-squared into the equation. Solve by the Square Root Method; note that x 2 = 4 has no solution in the Real Number System

2 4. Recall the format for the relationship between zeros and factors is: ( x k ) where k is a zero of the equation. So, expand (multiply) (x + 2)( x + 2)( x 1)(x 3) (x 2 + 4x + 4)(x 2 4x + 3) x 4 9x 2 4x +12 = f (x) If the form is ( x k ) then, for example (x ( 2)) = (x + 2) Multiply in any order that is convenient; Simplify 5. Use synthetic division because it explicitly;y asks for it! Remember to hold the place of any missing power of x using a x 2 + 2x + 3 Since remainder is zero, negative 2 divides f(x) evenly and is also a zero of f(x). That is, f( 2)=0 Since started with third degree and divided, then result must be quadratic of this form 6. There is no short-cut. The only way to know which, if any, are zeros for the given function is to actually do synthetic division using each of the given numbers. 2 / 3 does not work, but 1 and 4 both do.

3 7. If 2 is a zero, then the factor (x + 2) must evenly divide f(x). If so, then 2 should yield a zero when used for synthetic division x 2 3x 4 (x 4)( x +1) divides evenly since no remainder. Result factors to So, f (x ) = x 3 x 2 10x 8 = ( x 4)( x +1)(x + 2) Putting it all together shows f(x) factored completely into each of its linear factors 8. This is done the same way number 7 was done. Use synthetic division to reduce the degree and factor the resultant quadratic using conventional techniques. f (x) = x 3 + 4x 2 + x 6 = ( x + 2)( x + 3)( x 1) 9. Using synthetic division, the value of the remainder is the value of the function for the particular value of x used in the division This is the remainder According to the remainder theorem, the value of the remainder is the value of f(x) when x = 2. That is: f ( 2) = 28

4 10. Descartes Rule of Signs says that we can get some idea of the number of possible real solutions to an equation by counting the sign changes. Note that this does not tell us what the solutions are, only whether we can expect to find any real solutions or not. It works with the rule that says that complex imaginary solutions occur only in conjugate pairs. Remember, then, that the rule also states there may be that many zeros (equal to the number of sign changes) or less than that by an even number (duh! 2!) f (x) = 5x 4 3x 3 4x sign changes twice begins as positive, changes to negative, then changes to positive 2 sign changes means that there are 2 or 0 possible positive real zeros. Next, evaluate the function for ( x), simplify, determine the sign changes and, hence, the number of possible negative zeros. f ( x) = 5( x) 4 3( x) 3 4( x) + 2 f ( x) = 5x 4 + 3x 3 + 4x Evaluate and simplify. The sign doesn t change at all! There must not be any negative real zeros to this function. That is, there are no negative real numbers that will make the equation 0 = 5(x ) 4 3(x) 3 4(x) + 2 true. 11. Same problem, different function. There are 2 or 0 possible positive real zeros for this function; and there are 3 or 1 possible negative real zeros that will satisfy the equation = 0. This means there must be at least one real solution (a negative real number). f (x) = 6x 5 6x 3 +10x or 0 possible positive real zeros f ( x) = 6( x) 5 6( x) 3 +10( x) + 5 f ( x) = 6x 5 + 6x 3 10x sign changes so 3 or 1 possible negative real zeros

5 12. The rational roots test can give some indication of possible rational solutions to f(x) = 0. Use the factors of the constant term over the factors of the leading coefficient and make a comprehensive list from the result. Note that Descartes Rule may tell us there are real solutions, but they may not show up in this list since this covers only rational solutions. factorsof constant term factors of leadingcoefficient 2 3 ±1, ± 2 ±1, ± 3 ±1, ±2, ± 1 3, ± The question is find the solutions to the equation: 3x 4 27x 3 70x 2 = 0 To begin, note this is a fourth degree equation. You can use the algebra tools (see notes) to reduce the degree, if possible, and solve. Or, inspect the given equation and see if factoring is a possibility. Remember, the first rule of factoring is to first remove the GCF. 3x 4 27x 3 70x 2 = 0 x 2 ( 3x 2 27x 70) = 0 x 2 = 0 or 3x 2 27x 70 = 0 x = 0 or?????? Since 3x 2 27x 70 = 0 is not factorable, then the next best technique to solve is to use the Quadratic Formula. remove GCF and use principle of zero products. It sure would be nice if this quadratic would factor. At least it is a quadratic. 3x 2 27x 70 = 0 x = ( 27) ± ( 27) 2 4(3)( 70) 2(3) = 27 ± = 27 ± Use the Quadratic Formula. Note that this is not a pretty solution; sometimes, it s just that way. Not all problems have neat, simple solutions. This cannot be further reduced.

6 14. Find all the zeros means to solve the equation when f(x) = 0. Again, this is a higher degree equation; use the tools to reduce this to a quadratic which you know how to solve. By the rational roots test, possible solutions include: ±1, ± 1 2, ± 1 Now, I know that 4 most of you are not even going to try the fractions unless somebody makes you. So try +1or 1 using synthetic division to seek a zero which will also reduce the equation to a quadratic. It turns out that 1 is a solution! (yipee! the only way to find a possible solution is to use trial-anderror.) 4x 3 3x 1 = Okay. So 1 works and is a solution and (x 1) is a factor of f(x). Furthermore, the equation has been reduced to a quadratic which can be solved by the usual techniques. In fact, this will factor 4x 2 + 4x + 1 = 0 4x 2 + 2x + 2x + 1 = 0 ( ) +1( 2x +1) = 0 2x 2x +1 (2x +1)(2x +1) = 0 equate to zero and solve Using magic number technique and factor by grouping. x = 1 2 Yields the double root So, all the real zeros of the given function: 1 2,1

7 15. Write as a product of linear factors: x x There are a couple of ways to do this. (1) Note the quadratic-like expression and factor accordingly. x x (x 2 + 9)( x 2 +16) (x + 3i )(x 3i)( x + 4i)( x 4i) Factor the quadratic-like expression Irreducible over the reals, so use imaginary part of complex Linear factors have the variable of the 1st degree (2) Sometimes it s not easy to see how an expression can be factored. Remember that substitution is an excellent and useful technique ideal for these situations. x x u u (u + 9)(u +16) Use the substitution u = x 2 Factor. Then re-substitute and simplify. u = x 2 Write as linear factors. See the (x 2 + 9)( x 2 +16) difference of squares for technique. (x + 3i )(x 3i)( x + 4i)( x 4i) 16. Again, factor completely. This is an obvious difference of squares. Another difference of squares x 4 16 ( x 2 4) x ( ) ( x 2 4) ( x 2 + 4) (x 2)( x + 2)( x 2i)( x + 2i ) These factors are not linear! Requires use of imaginary part of complex numbers to factor as difference of squares. Factoring completely into linear factors.

8 17. Using the given zeros and the factor theorem. ( x zero) is a factor (x 1)(x +1)(x 0)( x 2) ( x 2 1)x( x 2) Expand. (multiply) in ( x 2 1) ( x 2 2x whatever way is convenient ) f (x ) = x 4 2x 3 x 2 + 2x format for using zeros as factors Note fourth degree function 18. Trick question. Note asks for fourth degree polynomial,but only three factors are given. Recall that complex solutions occur only in conjugate pairs. Therefore, 2i must also be a zero. ( x 1) x + 3 ( x 2 + 2x 3) x ( )( x 2i) ( x + 2i) ( ) f (x) = x 4 +2x 3 + x 2 + 8x 12 Write the linear factors and expand 19. Using the given information and the tools from this section, determine the zeros for the function. The idea is to reduce this higher degree polynomial to a quadratic which can then be solved in typical manner. Along the way, the other solutions will be exposed. f (x) = x 4 6x 3 +14x 2 54x + 45 Use synthetic division with the given zero i 3i 9 18i i i 5 18i 15i 3i 3i + 18i 15i Note that 3i must also be a zero and multiple synthetic division is okay; in fact, successive synthetic division works great! Now have quadratic to set equal to zero and solve. Don t forget the zeros found with synthetic division x 2 6x + 5 = 0 (x 5)(x 1) = 0 x = 5 or x = 1 { 1,5, ± 3i } Solve the resultant quadratic. Don t forget the solutions from the previous synthetic div

9 20. In order to factor this function, use the same techniques as in the previous problem. The factors and the zeros are closely related. Use any necessary tools to reduce the function to a quadratic, which you already know how to factor (and solve), and along the way you should discover the remaining solutions. Use synthetic division to reduce, but what numbers? Recall the Rational Roots test and, through trial-and-error, determine what numbers will be zeros. Again, successive synthetic division is possible. Using the Rational Roots Test, list the possible rational zeros. ±1, ± 2, ± 4, ± 8, ±16 ±1 ± 1, ± 2, ± 4, ± 8, ±16 are possible. Through trial-and-error, find one or more that leave a remainder of zero and thus are solutions to f(x) = 0 and are the corresponding factors. Use synthetic division x = ( x 2i) ( x + 2i) Using synthetic division and trial-anderror, find two solutions. (Note successive synthetic divisions). The final row yields a quadratic, which can be factored as the difference of squares Now putting everything together: linear factors of f(x). f (x) = ( x 2i) ( x + 2i) ( x 1) ( x 4) yields all the