Higher. Polynomials and Quadratics 64

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1 hsn.uk.net Higher Mathematics UNIT OUTCOME 1 Polnomials and Quadratics Contents Polnomials and Quadratics 64 1 Quadratics 64 The Discriminant 66 3 Completing the Square 67 4 Sketching Parabolas 70 5 Determining the Equation of a Parabola 7 6 Solving Quadratic Inequalities 74 7 Intersections of Lines and Parabolas 76 8 Polnomials 77 9 Snthetic Division Finding Unknown Coefficients 8 11 Finding Intersections of Curves 84 1 Determining the Equation of a Curve Approimating Roots 88 HSN100 This document was produced speciall for the HSN.uk.net website, and we require that an copies or derivative works attribute the work to Higher Still Notes. For more details about the copright on these notes, please see

2 Unit Polnomials and Quadratics OUTCOME 1 Polnomials and Quadratics 1 Quadratics A quadratic has the form numbers, provided a 0. You should alread be familiar with the following. a + b + c where a, b, and c are an real The graph of a quadratic is called a parabola. There are two possible shapes: concave up (if a > 0 ) concave down (if a < 0 ) This has a minimum turning point This has a maimum turning point To find the roots (i.e. solutions) of the quadratic equation we can use: factorisation; completing the square (see Section 3); the quadratic formula: EXAMPLES 1. Find the roots of ( + 1)( 3). Solve 3 a b c + +, b ± b = (this is not given in the eam). a or 3 = 1 = 3. ( )( ) or + 4 = 4 = 4. Page 64 HSN000

3 Unit Polnomials and Quadratics 3. Find the roots of We cannot factorise + 4 1, but we can use the quadratic formula: 4 ± ( 1) = 1 4 ± = 4 ± 0 = = ± = ± 5. Note If there are two distinct solutions, the curve intersects the -ais twice. If there is one repeated solution, the turning point lies on the -ais. If b < 0 when using the quadratic formula, there are no points where the curve intersects the -ais. Page 65 HSN000

4 Unit Polnomials and Quadratics The Discriminant Given a + b + c, we call b the discriminant. This is the part of the quadratic formula which determines the number of real roots of the equation a + b + c. If b If b > 0, the roots are real and unequal (distinct)., the roots are real and equal (i.e. a repeated root). If b < 0, the roots are not real; the parabola does not cross the -ais. EXAMPLE 1. Find the nature of the roots of a = 9 b = 4 c = 16 Since b b = = , the roots are real and equal.. Find the values of q such that Since a = 6 b = 1 c = q 6 1 q q = has real roots, b b q q q 4q 144 q = has real roots. 0 : two roots one root no real roots Page 66 HSN000

5 Unit Polnomials and Quadratics 3. Find the range of values of k for which the equation no real roots. k + 7 has For no real roots, we need a = k b = c = 7 b b < 0 : < 0 4 k ( 7) < k < 0 8k < 4 4 k < 8 k <. 4. Show that ( k ) ( k ) ( k ) has real roots for all real values of k. a = k + 4 b = 3k + c = k b 1 7 ( 3k ) 4( k 4)( k ) 9k 1k 4 ( k 4)( 4k 8) = + + = = k + k + k k = + + ( k ) 1k 36 = + 6. Since b ac ( k ) 4 = + 6 0, the roots are alwas real. 3 Completing the Square The process of writing a b c called completing the square. = + + in the form ( ) = a + p + q is Once in completed square form we can determine the turning point of an parabola, including those with no real roots. The ais of smmetr is = p and the turning point is ( p, q). + p = + p + p. For eample, The process relies on the fact that ( ) we can write the epression + 4 using the bracket ( + ) since when multiplied out this gives the terms we want with an etra constant term. Page 67 HSN000

6 Unit Polnomials and Quadratics This means we can rewrite the epression gives us the correct and terms, with an etra constant. + k using ( + k ) since this We will use this to help complete the square for = Step 1 Make sure the equation is in the form = a + b + c. Step Take out the -coefficient as a factor of the and terms. Step 3 Replace the + k epression and compensate for the etra constant. = ( ) = = 3( ( + ) 4) 3 = 3( + ) 1 3. Step 4 Collect together the constant terms. = 3( + ) 15. Now that we have completed the square, we can see that the parabola with equation = has turning point (, 15). EXAMPLES 1. Write = in the form = ( + p) + q. = = ( + 3) 9 5 = ( + 3) 14. Note You can alwas check our answer b epanding the brackets.. Write ( ) ( 3 ) in the form ( + p) + q. = = Page 68 HSN000

7 Unit Polnomials and Quadratics 3. Write = in the form = ( + a) + b and then state: (i) the ais of smmetr, and (ii) the minimum turning point of the parabola with this equation. = ( ) ( ) = = (i) The ais of smmetr is = 4. (ii) The minimum turning point is ( 4, 19). 4. A parabola has equation = (a) Epress the equation in the form = ( + a) + b. (b) State the turning point of the parabola and its nature. (a) = ( ) = (( 3 9 ) 4 ) ( 3 ) 3 ( ) = = = 4. (b) The turning point is 3 ( ), and is a minimum. Remember If the coefficient of is positive then the parabola is concave up. Page 69 HSN000

8 Unit Polnomials and Quadratics 4 Sketching Parabolas The method used to sketch the curve with equation depends on how man times the curve intersects the -ais. = a + b + c We have met curve sketching before. However, when sketching parabolas, we do not need to use calculus. We know there is onl one turning point, and we have methods for finding it. Parabolas with one or two roots Find the -ais intercepts b factorising or using the quadratic formula. Find the -ais intercept (i.e. where ). The turning point is on the ais of smmetr: O The ais of smmetr is halfwa between two distinct roots. O A repeated root lies on the ais of smmetr. Parabolas with no real roots There are no -ais intercepts. Find the -ais intercept (i.e. where ). Find the turning point b completing the square. EXAMPLES 1. Sketch the graph of = Since b 4 ac = ( 8 ) > 0, the parabola crosses the -ais twice. The -ais intercept ( ) : = ( 0) 8( 0) + 7 = 7 ( 0, 7 ). The -ais intercepts ( ) : ( 1)( 7) 1 or 7 = 1 = 7 ( 1, 0 ) ( 7, 0 ). The ais of smmetr lies halfwa between = 1 and = 7, i.e. = 4, so the -coordinate of the turning point is 4. Page 70 HSN000

9 Unit Polnomials and Quadratics We can now find the -coordinate: ( ) ( ) = = = 9. So the turning point is ( 4, 9).. Sketch the parabola with equation Since b 4ac ( 6) 4 ( 1) ( 9) 0 = 6 9. = =, there is a repeated root. The -ais intercept ( ) : = ( 0) 6( 0) 9 = 9 ( 0, 9 ). Since there is a repeated root, ( 3, 0) is the turning point. 3. Sketch the curve with equation The -ais intercept ( ) : = ( ) ( + 3)( + 3) + 3 = 3 = ( 3, 0 ). Since b 4 ac = ( 8 ) 4 13 < 0, there are no real roots. The -ais intercept ( ) : = ( 0) 8( 0) + 13 = 13 ( 0,13 ). So the turning point is (, 5 ). Complete the square: = O 3 9 = 6 9 ( ) = = ( ) O = ( ) + 5. = + = ( 4, 9) 8 13 O (, 5) Page 71 HSN000

10 Unit Polnomials and Quadratics 5 Determining the Equation of a Parabola Given the equation of a parabola, we have seen how to sketch its graph. We will now consider the opposite problem: finding an equation for a parabola based on information about its graph. We can find the equation given: the roots and another point, or the turning point and another point. When we know the roots If a parabola has roots where k is some constant. = a and = b then its equation is of the form = k ( a)( b) If we know another point on the parabola, then we can find the value of k. EXAMPLES 1. A parabola passes through the points ( 1, 0 ), ( 5, 0 ) and ( 0, 3 ). Find the equation of the parabola. Since the parabola cuts the -ais where = 1 and = 5, the equation is of the form: = k ( 1)( 5 ). To find k, we use the point ( 0, 3 ): = k ( 1)( 5) 3 = k ( 0 1)( 0 5) 3 = 5k k = 3 5. So the equation of the parabola is: = ( 1)( 5) ( 6 5) = = Page 7 HSN000

11 Unit Polnomials and Quadratics. Find the equation of the parabola shown below. ( 1, 6) Since there is a repeated root, the equation is of the form: = k ( + 5)( + 5) = k ( + 5 ) Hence = ( + ). 5 O To find k, we use ( 1, 6 ) : = k ( + 5) 6 = k ( 1+ 5) k = 6 =. When we know the turning point Recall from Completing the Square that a parabola with turning point p, q has an equation of the form ( ) where a is some constant. = a( + p) + q If we know another point on the parabola, then we can find the value of a. EXAMPLE 3. Find the equation of the parabola shown below. O ( 4, ) 7 Since the turning point is ( 4, ), the equation is of the form: = a ( 4). = Hence ( ) To find a, we use ( 0, 7) : ( ) = a 4 ( ) 7 = a a = 5 a = Page 73 HSN000

12 Unit Polnomials and Quadratics 6 Solving Quadratic Inequalities The most efficient wa of solving a quadratic inequalit is b making a rough sketch of the parabola. To do this we need to know: the shape concave up or concave down, the -ais intercepts. We can then solve the quadratic inequalit b inspection of the sketch. EXAMPLES 1. Solve + 1 < 0. The parabola with equation The -ais intercepts are given b: + 1 ( + 4)( 3) + 4 = 4 or 3 = 3. = + 1 is concave up. Make a sketch: 4 = So + 1 < 0 for 4 < < 3.. Find the values of for which The parabola with equation The -ais intercepts are given b: ( ) ( 3 + )( 3) 3 + or 3 = = 3. = is concave down. Make a sketch: = So for 3 3. Page 74 HSN000

13 Unit Polnomials and Quadratics 3. Solve > 0. The parabola with equation = is concave up. The -ais intercepts are given b: Make a sketch: ( 1)( 3) = or 3 = 1 = So > 0 for < 1 and > Find the range of values of for which the curve is strictl increasing. We have d 4 5 d = +. The curve is strictl increasing where > Make a sketch: ( 1)( + 5) 1 or + 5 = 1 = = Remember Strictl increasing means d 0 d >. = So the curve is strictl increasing for < 5 and > Find the values of q for which ( ) For no real roots, b < 0 : a = 1 b = q q b = q 4 = ( q 4)( q 4) q c = 1 q = q 8q + 16 q + q q has no real roots. ( ) ( )( ) q 10q 16. = + We now need to solve the inequalit q 10q + 16 < 0. The parabola with equation q q = is concave up. Page 75 HSN000

14 Unit Polnomials and Quadratics The -ais intercepts are given b: q 10q + 16 ( q )( q ) 8 q or q 8 Therefore b q = q = 8. < 0 for q 8 no real roots when < q < 8. Make a sketch: = q 10q + 16 < <, and so ( ) 7 Intersections of Lines and Parabolas 8 + q q has To determine how man times a line intersects a parabola, we substitute the equation of the line into the equation of the parabola. We can then use the discriminant, or factorisation, to find the number of intersections. q If b > 0, the line and curve intersect twice. If b, the line and curve intersect once (i.e. the line is a tangent to the curve). If b < 0, the line and the parabola do not intersect. EXAMPLES 1. Show that the line = 5 is a tangent to the parabola and find the point of contact. Substitute = 5 into: = + 5 = ( 1)( 1). Since there is a repeated root, the line is a tangent at = 1. = + To find the -coordinate, substitute = 1 into the equation of the line: = 5 1 = 3. So the point of contact is ( 1, 3 ). Page 76 HSN000

15 Unit Polnomials and Quadratics. Find the equation of the tangent to = + 1 that has gradient 3. The equation of the tangent is of the form = m + c, with m = 3, i.e. = 3 + c. Substitute this into c = c = Since the line is a tangent: b ( 3) 4 ( 1 c ) c 4c = 5 c = 5 4. Therefore the equation of the tangent is: = Note You could also do this question using methods from Differentiation. 8 Polnomials Polnomials are epressions with one or more terms added together, where each term has a number (called the coefficient) followed b a variable (such as ) raised to a whole number power. For eample: or The degree of the polnomial is the value of its highest power, for eample: has degree has degree 18. Note that quadratics are polnomials of degree two. Also, constants are 0 polnomials of degree zero (e.g. 6 is a polnomial, since 6 = 6 ). Page 77 HSN000

16 Unit Polnomials and Quadratics 9 Snthetic Division Snthetic division provides a quick wa of evaluating polnomials. 3 For eample, consider f ( ) = Evaluating directl, we find f ( 6) = 11. We can also evaluate this using snthetic division with detached coefficients. Step 1 Detach the coefficients, and write them across the top row of the table. Note that the must be in order of decreasing degree. If there is no term of a specific degree, then zero is its coefficient. Step Write the number for which ou want to evaluate the polnomial (the input number) to the left Step 3 Bring down the first coefficient Step 4 Multipl this b the input number, writing the result underneath the net coefficient. Step 5 Add the numbers in this column. Repeat Steps 4 and 5 until the last column has been completed. The number in the lower-right cell is the value of the polnomial for the input value, often referred to as the remainder = f ( 6) Page 78 HSN000

17 Unit Polnomials and Quadratics EXAMPLE 3 1. Given f ( ) = + 40, evaluate f ( ) using snthetic division So f ( ). using the above process Note In this eample, the remainder is zero, so f ( ). 3 This means + 40 when =, which means that = is a root of the equation. So + must be a factor of the cubic. We can use this to help with factorisation: ( ) f ( ) = ( + ) q ( ) where q ( ) is a quadratic Is it possible to find the quadratic q ( ) using the table? Tring the numbers from the bottom row as coefficients, we find: ( + )( 0) 3 = + 3 = 40 = f ( ) So using the numbers from the bottom row as coefficients has given the correct quadratic. In fact, this method alwas gives the correct quadratic, making snthetic division a useful tool for factorising polnomials. EXAMPLES. Show that 4 is a factor of 4 is a factor = 4 is a root Since the remainder is zero, = 4 is a root, so 4 is a factor. Page 79 HSN000

18 Unit Polnomials and Quadratics 3 3. Given f ( ) = , show that = 7 is a root of f ( ), and hence full factorise f ( ) Since the remainder is zero, = 7 is a root. 3 Hence we have f ( ) = Show that = 5 is a root of factorise the cubic ( 7)( 7 1) = + + Using snthetic division to factorise ( )( )( ) = =, and hence full Since = 5 is a root, + 5 is a factor. ( )( ) = This does not factorise an further since the quadratic has b < 0. In the eamples above, we have been given a root or factor to help factorise polnomials. However, we can still use snthetic division if we do not know a factor or root. Provided that the polnomial has an integer root, it will divide the constant term eactl. So b tring snthetic division with all divisors of the constant term, we will eventuall find the integer root. 5. Full factorise Numbers which divide 15: ± 1, ± 3, ± 5, ± Tr = 1: ( 1) + 5( 1) 8( 1) 15 = Tr = 1: ( 1) + 5( 1) 8( 1) 15 = Note For ±1, it is simpler just to evaluate the polnomial directl, to see if these values are roots. Page 80 HSN000

19 Unit Polnomials and Quadratics Tr = 3: Since = 3 is a root, 3 is a factor. 3 So = ( 3)( ) = ( 3)( + 1)( + 5 ). Using snthetic division to solve equations We can also use snthetic division to help solve equations. EXAMPLE 6. Find the solutions of =. Numbers which divide 1: ± 1, ±, ± 3, ± 4, ± 6, ± 1. 3 Tr = 1: ( 1) 15( 1) + 16( 1) + 1 = Tr = 1: ( 1) 15( 1) + 16( 1) + 1 Tr = : = Since = is a root, is a factor: = The Factor Theorem and Remainder Theorem For a polnomial f ( ): If f ( ) is divided b h or + 1 = 1 then the remainder is f ( ) f ( h) h is a factor of f ( ). h, and As we saw, snthetic division helps us to write f ( ) in the form where ( ) ( h) q ( ) + f ( h) q is called the quotient and f ( ) ( )( ) 11 6 ( )( + 1)( 6) h the remainder. or 6 = 6. Page 81 HSN000

20 Unit Polnomials and Quadratics EXAMPLE 3 7. Find the quotient and remainder when f ( ) = is divided b + 1, and epress ( ) The quotient is 4 3 ( ) ( )( ) f = f as ( 1) q( ) f ( h) and the remainder is 3, so 10 Finding Unknown Coefficients Consider a polnomial with some unknown coefficients, such as 3 + p p + 4, where p is a constant. If we divide the polnomial b h, then we will obtain an epression for the remainder in terms of the unknown constants. If we alread know the value of the remainder, we can solve for the unknown constants. EXAMPLES 1. Given that 3 is a factor of 3 is a factor = 3 is a root p p p p 3 p Since = 3 is a root, the remainder is zero: p 3 p = 4 p = , find the value of p. Note This is just the same snthetic division procedure we are used to. Page 8 HSN000

21 Unit Polnomials and Quadratics 3. When f ( ) = p + q q is divided b, the remainder is 6, and 1 is a factor of f ( ). Find the values of p and q. When f ( ) is divided b, the remainder is 6. p q 17 4q p 4 p + q 8 p + 4q 34 p p + q 4 p + q 17 8p + 8q 34 Since the remainder is 6, we have: 8 p + 8q 34 = 6 8 p + 8q = 40 p + q = 5. 1 Since 1 is a factor, f ( 1) : 3 f ( 1) = p( 1) + q ( 1) 17( 1) + 4q = p + q q = p + 5q 17 i.e. p + 5q = 17. Note There is no need to use snthetic division here, but ou could if ou wish. Solving 1 and simultaneousl, we obtain: 1: 4q = 1 q = 3. Put q = 3 into 1: p + 3 = 5 Hence p = and q = 3. p =. Page 83 HSN000

22 Unit Polnomials and Quadratics 11 Finding Intersections of Curves We have alread met intersections of lines and parabolas in this outcome, but we were mainl interested in finding equations of tangents We will now look at how to find the actual points of intersection and not just for lines and parabolas; the technique works for an polnomials. EXAMPLES 1. Find the points of intersection of the line = 4 4 and the parabola = 1. To find intersections, equate: 1 = ( )( ) = 1 or = 4. Find the -coordinates b putting the -values into one of the equations: when = 1, = 4 ( 1) 4 = 4 4 = 8, when = 4, = = 16 4 = 1. So the points of intersection are ( 1, 8) and ( 4,1 ).. Find the coordinates of the points of intersection of the cubic 3 = and the line = To find intersections, equate: = ( )( ) ( 1)( 3)( 5) = 1 or = 3 or = 5. Find the -coordinates b putting the -values into one of the equations: when = 1, = = =, when = 3, = = = 4, when = 5, = = = 10. So the points of intersection are ( ) 1,, ( 3, 4) and ( 5, 10). Remember You can use snthetic division to help with factorising. Page 84 HSN000

23 Unit Polnomials and Quadratics 3. The curves = + 4 and 3 = are shown below. 3 = A O B C = + 4 Find the -coordinates of A, B and C, where the curves intersect. To find intersections, equate: 3 + = + 3 ( )( ) ( )( )( ) = 1 or = or = 4. Remember You can use snthetic division to help with factorising. So at A, = 1; at B, = ; and at C, = Find the -coordinates of the points where the curves 3 and = intersect. = To find intersections, equate: = ( )( ) ( + 1)( 3)( 5) = 1 or = 3 or = 5. So the curves intersect where = 1,3,5. Page 85 HSN000

24 Unit Polnomials and Quadratics 1 Determining the Equation of a Curve Given the roots, and at least one other point ling on the curve, we can establish its equation using a process similar to that used when finding the equation of a parabola. EXAMPLE 1. Find the equation of the cubic shown in the diagram below. 6 3 O 1 36 Step 1 Write out the roots, then rearrange to get the factors. Step The equation then has these factors multiplied together with a constant, k. Step 3 Substitute the coordinates of a known point into this equation to find the value of k. Step 4 Replace k with this value in the equation. = 6 = 3 = = k ( + 6)( + 3)( 1 ). Using ( 0, 36) : k ( 0 + 6)( 0 + 3)( 0 1) = 36 k ( 3)( 1)( 6) = 36 18k = 36 = ( + 6)( + 3)( 1) k =. ( )( ) 3 ( ) = = = Page 86 HSN000

25 Unit Polnomials and Quadratics Repeated Roots If a repeated root eists, then a stationar point lies on the -ais. Recall that a repeated root eists when two roots, and hence two factors, are equal. EXAMPLE. Find the equation of the cubic shown in the diagram below. 9 O 3 = = 3 = So = k( + )( 3). Use ( 0, 9 ) to find k : 9 = k ( 0 + )( 0 3) 9 = k 9 k = 1. Note = 3 is a repeated root, so the factor ( 3) appears twice in the equation. 1 = = = So = ( + )( 3) ( )( 6 9) 3 ( ) 1 3 Page 87 HSN000

26 Unit Polnomials and Quadratics 13 Approimating Roots Polnomials have the special propert that if f ( a ) is positive and f ( b ) is negative then f must have a root between a and b. f ( a) a b f ( b) We can use this propert to find approimations for roots of polnomials to an degree of accurac b repeatedl zooming in on the root. EXAMPLE 3 Given f ( ) = 4 + 7, show that there is a real root between = 1 and =. Find this root correct to two decimal places. Evaluate f ( ) at = 1 and = : 3 f ( 1) = 1 4( 1) ( 1) + 7 = 3 f ( ) = 4( ) ( ) + 7 = 5 Since f ( 1) > 0 and f ( ) < 0, f ( ) has a root between these values. Start halfwa between = 1 and =, then take little steps to find a change in sign: f ( 1. 5) = < 0 f ( 1. 4) = < 0 f ( 1. 3) = < 0 f ( 1. ). 568 > 0. Since f ( 1. ) > 0 and f ( 1. 3) < 0, the root is between = 1. and = Start halfwa between = 1. and = 1. 3 : f ( 1. 5) > 0 f ( 1. 6) > 0 f ( 1. 7) > 0 f ( 1. 8) = < 0. Since f ( 1. 7) > 0 and f ( 1. 8) < 0, the root is between these values. Finall, f ( 1.75) > 0. Since f ( 1. 75) > 0 and f ( 1. 8) < 0, the root is between = and = Therefore the root is = 1. 8 to d.p. 1 Page 88 HSN000