Polynomials. Teachers Teaching with Technology. Scotland T 3. Teachers Teaching with Technology (Scotland)

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1 Teachers Teaching with Technology (Scotland) Teachers Teaching with Technology T Scotland Polynomials Teachers Teaching with Technology (Scotland)

2 POLYNOMIALS Aim To demonstrate how the TI-8 can be used to facilitate a fuller understanding of polynomials and show clearly the relationship between the algebraic solution and the graphical solution. Objectives Mathematical objectives By the end of this unit you should know: how torecognise features of various polynomials the relationship between the graph of a sitution and the algebra used to describe it the relationship between roots and factors how to factorise and solve polynomial equations how to find approimate roots of a polynomial by decimal search how to find polynomial coefficients Calculator objectives By the end of this unit you should be able to draw graphs using [Y=] alter the display of a graph using [WINDOW] and [ZOOM]. use the [nd][table] function with appropriate setting, using [nd][tblset] : T Scotland Polynomials Page 1 of 9

3 Factorising and Solving Polynomial Equations Calculator skills sheet Using a TI - 8 to assist you in solving polynomials not only reduces the chances of you making a silly error but also makes the whole process much faster. It can also help you gain a fuller understanding of the mathematics. Here is a typical question from a tetbook and a method for solution. Find the roots of = Before going any further check that the [MODE] screen looks like this, particularly ensure that the TI is not rounding answers by highlighting Float. Rounded answers could appear as non-eistent roots. Enter the function on the [Y=] screen and graph the result on the [ZOOM 6:ZStandard] window range. For some functions [ZOOM 4:ZDecimal] is appropriate. The graph of this function shows that it has roots at values which look like -1, and. This can be confirmed by looking at a table of values over this range. Using the [nd][tblset] screen and then [nd][table] we can see that the values of the function are indeed zero at -1, and. Hence we can now say that = -1, = and = are roots. Once we have obtained the roots from the calculator it is easy to say what the factors of the equation are and so this same method can be used to fully factorise a polynomial. The process to be used is almost identical. : T Scotland Polynomials Page of 9

4 Eample 1 Find the roots of = WORKED SOLUTION From graphic calculator root at = No remainder = -1 is a root and ( + 1) is a factor. (Factor Theorem) 5+ 6 This quadratic must factorise to give ( - ) and ( - ). We already know this from the graph and table but can confirm it by multiplying out the bracket or by factorising the quadratic = ( + 1)( ) = = ( + 1)( - )( - ) = + 1 = or - = or - = = - 1 or = or = Eercise 1 Solve these Polynomial Equations = = = = = : T Scotland Polynomials Page of 9

5 Eample Fully factorise This function is a cubic (polynomial of degree ) which means it may have: Roots Factors i.e = a, = b, = c Unique Linear Factors ( - a)( - b)( - c) i.e = a, = b, = b 1 Unique Linear Factor & 1 Repeated pair ( - a)( - b)( - b) 1 i.e = a 1 Linear Factor & 1 Non-factorising Quadratic [ZOOM 6:ZStandard] From the TI it can be seen that this function has 1 unique root and 1 pair of coincidental roots, i.e. it must have 1 unique linear factor and 1 pair of repeated linear factors. The table display shows that the roots are = -1 and = 1, from the graph we can see that the roots at = 1, are coincidental. The roots of this function are = -1, = 1 and = 1. The factors must be ( + 1) and ( - 1) and ( - 1) WORKED SOLUTION From graphic calculator root at = No remainder = -1 is a root and ( + 1) is a factor. (Factor Theorem) This quadratic must factorise to give ( - 1) and ( - 1) = ( + 1)( - + 1) = ( + 1)( - 1)( - 1) : T Scotland Polynomials Page 4 of 9

6 Eample Fully factorise and solve [ZOOM 4:ZDecimal] = From the TI it can be seen that this function has unique roots, i.e. it must have linear factors. From the graph we can see that one of the roots is fractional (about =.5). The table display confirms this showing roots at = -, = -1 and another between = and = 1, since f() = - and f(1) = 8. The roots of this function are = -, = -1 and =.5 The factors must be ( + ) and ( + 1) and ( -.5) WORKED SOLUTION From graphic calculator root at = - By changing the [nd][tbl SET] as shown we can see that the fractional root is indeed at = No remainder = - is a root and ( + ) is a factor. (Factor Theorem) + 1 This quadratic must factorise to give ( + 1) and ( -.5) We do not quote linear factors with fractional values so we can multiply the factor to eliminate the fractional value, in this case by so ( -.5) = ( - 1) Hence the factors of the resulting quadratic are: ( + 1) and ( - 1) = ( + )( - - 1) = ( + )( + 1)( - 1) + = or + 1 = or - 1 = = - or = - 1 or = 1 1 = - or = - 1 or = : T Scotland Polynomials Page 5 of 9

7 Approimate Roots of a Polynomial (Decimal Search) Eample 1 Show that there is a real root of the equation between 1 and, and find an approimation for the root correct to decimal places. + = On your TI enter the function and draw the graph. [ZOOM 4:ZDecimal] From this graph it can be seen that the function does have a root between = 1 and = Using [nd][tbl SET] set up a table of values as shown. The resulting table shows that the root does lies between = 1 and = Using [nd][tbl SET] change the starting value of the table and the step size. The resulting table shows that the root does lies between = 1. and = 1. Change the start and step size again. The resulting table shows that the root does lies between = 1.1 and = 1. Change the start and step size once more. The resulting table shows that the root does lies between = 1.1 and = so to d.p. the root is = 1.1 WORKING f ( ) = + f () 1 = 1 f ( ) = 7 f (.) 1 = 7. f (.) 1 = 497. f (. 11) =. 184 f (. 1) =. 585 f (. 11) =. f (. 114) =. 19 } } } } Root between 1 and, 1 < < Root between 1. and 1., 1. < < 1. Root between 1.1 and 1., 1.1 < < 1. Root is 1. to 1 d.p Root between 1.1 and 1.14, 1.1 < < 1.14 Root is 1.1 to d.p : T Scotland Polynomials Page 6 of 9

8 Finding Polynomial Coefficients Eample 1 If ( + ) is a factor of p , find the value of p. On your TI enter the function using the letter P as the coefficient of on the [Y=] screen. Note to obtain the letter P use [ALPHA][P] The TI will evaluate the function for P= the value stored as P in the memory of the calculator. Store the value zero as P. [][STO][ALPHA][P][ENTER] Using [nd][tblset] and [nd][table] obtain a value of this function when = -, the root which would accompany the factor ( + ). The value of the function when P = and = - is -7. Therefore the term P must equal 7, to make the value at = - be If P = 7, when = - then P = This answer can now be checked by storing the value as P. [][STO][ALPHA][P][ENTER] The table of results now shows that = - is a root and therefore ( + ) is a factor. WORKING Let so f ( ) = P g()= f ( ) = g() + P ( ) 4 4 f ( ) has a factor + f ( ) has a root at = f ( ) = g( ) + P( ) = but g( ) = 7 f ( ) = 7+ 9P = 9P = 7 P = : T Scotland Polynomials Page 7 of 9

9 Eercise (Like Eercise 1 only trickier) Solve these Polynomial Equations = = = 19+ = = Eercise = Eercise Fully factorise these polynomials Show that = has a root between and 1. Find the value of this root to 1 d.p.. Show that = has a root between and 1, and another between -1 and. Find the values of these roots to 1 d.p.. Show that = has a root between 1.5 and. Find the value of this root to d.p. 4. Find all of the roots of = to 1 d.p. 5. Find all of the roots of = to 1 d.p. 6. Find all of the roots of = to 1 d.p. Eercise 5 1. Given that + is a factor of - + k + 6 = Find the value of k.. Given that + 1 is a factor of a = Find the value of a.. Given that + 4 is a factor of p = Find the value of p and solve the equation. 4. Given that + - is a factor of p + q = Find the value of p and q. Hence factorise this equation fully. (HINT: Try to form a pair of simultaneous equations) : T Scotland Polynomials Page 8 of 9

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