11.Remainder and Factor Theorem

11: THE REMAINDER AND FACTOR THEOREM

Solving and simplifying polynomials Thus, from arithmetic, we know that we can express a In our study of quadratics, one of the methods used to dividend as: simplify and solve was factorisation. For example, we may solve for x in the following equation as follows: Dividend = (Quotient ´ Divisor) + Remainder 2 x + 5x + 6 = 0 When there is no remainder, � = 0, and the divisor is (x + 3)(x + 2) = 0 now a factor of the number, so x + 3 = 0,⇒ x = −3

x + 2 = 0 ⇒ x = −2 Dividend = Quotient ´ Factor

Hence, x = −3 or −2 are solutions or roots of the The process we followed in arithmetic when dividing quadratic equation. is very similar to what is to be done in algebra. Examine the following division problems in algebra A more general name for a quadratic is a polynomial and note the similarities. of degree 2, since the highest power of the unknown is two. The method of factorisation worked for Division of a polynomial by a linear expression quadratics whose solutions are integers or rational We can apply the same principles in arithmetic to numbers. dividing algebraic expressions. Let the quadratic For a polynomial of order 3, such as function �(�) represent the dividend, and (� − 1) the x3 + 4x2 + x − 6 = 0 the method of factorisation may divisor, where also be applied. However, obtaining the factors is not �(�) = 3� − � + 2 as simple as it was for quadratics. We would likely have to write down three linear factors, which may Quotient 3x + 2 prove difficult. In this section, we will learn to use the 2 Dividend remainder and factor theorems to factorise and to solve x −1 3x − x + 2 polynomials that are of degree higher than 2. − (3x2 − 3x)

Divisor 2x 2 + Before doing so, let us review the meaning of basic − (2x − 2) terms in division. 4 Remainder Terms in division We are familiar with division in arithmetic.

From the above example, we can deduce that: 3� − � + 2 = (3� + 2)(� − 1) + 4 Quotient 15 ↑ ↑ ↑ ↑ Dividend 32 489 Dividend Quotient Divisor Remainder 32 Divisor 169 Consider (� − 1) = 0, 160 �(�) = (3� + 2) × 0 + 4 = 4 9 Remainder But, (� − 1) = 0 implies that � = 1 The number that is to be divided is called the dividend. Thewww.faspassmaths.com dividend is divided by the divisor. Therefore, when � = 1, �(�) = 4 or �(1) = 4. The result is the quotient and the remainder is what is left over. We can conclude that when the polynomial 3� − � + 2 From the above example, we can deduce that: is divided by (� − 1), the remainder is 489 = (15 × 32) + 9 �(1) = 4 ↑ ↑ ↑ ↑ Dividend Quotient Divisor Remainder We shall now perform division using a cubic polynomial as our dividend.

Copyright © 2019. Some Rights Reserved. www.faspassmaths.com Pg 1 of 7 2 We can conclude that when the polynomial x +14x + 25 3 2 2� − 3� + 4� + 5 x − 2 x +12x − 3x + 4 is divided by (� + 2), the remainder is − (x3 − 2x2 ) �(−2) = −31 14x2 − 3x + 4 − (14x − 28x) The Remainder Theorem for divisor (� − �) 25x + 4 From the above examples, we saw that a polynomial −(25x − 50) can be expressed as a product of the quotient and the divisor plus the remainder: 54 3� − � + 2 = (3� + 2)(� − 1) + 4

From the above example, we can deduce that: � + 12� − 3� + 4 = (� + 14� + 25)(� − 2) + 54 � + 12� − 3� + 4 = (� + 14� + 25)(� − 2) + 54 ↑ ↑ ↑ ↑ 2� − 3� + 4� + 5 = (2� + � + 18)(� + 2) − 31 Dividend Quotient Divisor Remainder

We can now formulate the following expression Consider (� − 2) = 0, f (x) Q(x) �(�) = (� + 14� + 25) × 0 + 54 = 54 where, is a polynomial whose quotient is and whose remainder is R when divided by (x − a) . But, (� − 2) = 0 implies that � = 2 fx()= Qx ()´- ( x a )+ R

Therefore, when � = 2, �(�) = 54 or �(2) = 54. If we were to substitute � = � in the above expression, We can conclude that when the polynomial then our result will be equal to �, the remainder when � + 12� − 3� + 4 the divisor, (� − �) is divided by the polynomial, is divided by (� − 2), the remainder is �(�). �(2) = 54 We are now able to state the remainder theorem. Now consider another example of a cubic polynomial divided by a linear divisor. The Remainder Theorem If is any polynomial and is divided f (x) f (x) 2 by , then the remainder is 2x + 7x +18 (x − a) fa(). x + 2 2x3 − 3x2 + 4x + 5 The validity of this theorem can be tested in any of − (2x3 + 4x2 ) the equations above, for example: − 7x2 + 4x + 5 − (7x2 −14x) 1. When 3� − � + 2 was divided by (� − 1), the remainder was 4. 18x + 5 −(18x + 36) According to the remainder theorem, the remainder − 31 can be computed by substituting � = 1 in �(�) �(�) = 3� − � + 2 �(1) = 3(1) − 1 + 2 From the above example, we can deduce that: �(1) = 4 2� − 3� + 4�www.faspassmaths.com+ 5 = (2� + � + 18)(� + 2) − 31 ↑ ↑ ↑ ↑ 2. When � + 12� − 3� + 4 was divided by Dividend Quotient Divisor Remainder (� − 2), the remainder was 54.

Consider (� + 2) = 0, According to the remainder theorem, the remainder �(�) = (2� + � + 18) × 0 − 31 = −31 can be computed by substituting � = 2 in �(�)

�(�) = � + 12� − 3� + 4 But, (� + 2) = 0 implies that � = −2 �(2) = (2) + 12(2) − 3(2) + 4 Therefore, when � = −2, �(�) = −31 or �(2) = 8 + 48 − 6 + 4 �(−2) = −31. �(2) = 54

Copyright © 2019. Some Rights Reserved. www.faspassmaths.com Pg 2 of 7 The above rule is called the Factor Theorem, it is a 3. When 2� − 3� + 4� + 5 was divided by special case of the Remainder Theorem, when � = 0. (� + 2) the remainder was −31. The validity of this theorem can be tested by substituting � = 1 in each of the above functions. According to the remainder theorem, the remainder can be computed by substituting � = −2 in �(�) �(�) = 3� − � − 2 �(�) = 8� − 10� − � + 3

�(�) = 2� − 3� + 4� + 5 �(1) = 3(1) − 1 − 2 �(1) = 8(1) − 10(1) − 1 + 3 �(−2) = 2(−2) − 3(−2) + 4(−2) + 5 �(1) = 0 �(−2) = −16 − 12 − 8 + 5 �(1) = 0 �(−2) = −31

Example 1 In the above examples, when we let (� − 1) = 0, or Find the remainder when � = 1, �(�) = 0 because the remainder, � = 0. �(�) = 3� + � − 4� − 1 is divided by (� − 2).

The Factor Theorem If is any polynomial and is divided Solution f (x) f (x) by (x − a) , and the remainderfa()= 0 then By the Remainder Theorem, the remainder is �(2). (x − a) is a factor of f (x) �(�) = 3� + � − 4� − 1 �(2) = 3(2) + (2) − 4(2) − 1 �(2) = 24 + 4 − 8 − 1 Example 2 �(2) = 19 Show that (� − 2) is a factor of Hence, the remainder is 19 �(�) = 3� + � − 14�

The Factor Theorem for divisor (� − �) Solution Now, consider the following examples when there is no remainder. By the Factor Theorem, if (� − 2) is a factor of the remainder is zero. We now compute the 2 remainder, �(2). 3x + 2 8x − 2x − 3 �(�) = 3� + � − 14� 2 x −1 3x − x − 2 x −1 8x3 −10x2 − x + 3 �(2) = 3(2) + (2) − 14(2) − (3x2 − 3x) (8x3 8x2 ) �(2) = 24 + 4 − 28 − − �(2) = 0 2x 2 2 − − 2x − x + 3 (2x 2) 2 − − − (−2x + 2x) Hence, (� − 2) is a factor of �(�). 0 − 3x + 3 −(3x + 3) 0 The Remainder and Factor Theorem for divisor (�� + �) When the divisor is not in the form, (x − a) , but in the We can express the dividend as a product of the divisor general linear form (�� + �), the remainder can no and the quotientwww.faspassmaths.com only, since � = 0. longer be �(�). This is because the coefficient of x is not equal to one. �(�) = 3� − � − 2 = (3� + 2)(� − 1) �(�) = 8� − 10� − � + 3 = (8� − 2� − 3)(� − 1) Consider the following example, where the divisor is of the form, (�� + �). We can now formulate the following expression where �(�) is a polynomial, �(�) is the quotient and (� − �) Let (2� + 3) be a divisor of is a factor of the polynomial. �(�) = 2� + 7� + 2� + 9 We perform the division as shown below and note �(�) = �(�) × (� − �) that the remainder is 15.

Copyright © 2019. Some Rights Reserved. www.faspassmaths.com Pg 3 of 7 2 x + 2x − 2 where, f (x) is a polynomial whose quotient is Q(x) 2x + 3 2x3 + 7x2 + 2x + 9 and the remainder is R when divided by (�� + �). 3 2 By setting (�� + �) = 0, the above polynomial will − (2x + 3x ) 2 become 4x + 2x + 9 ( ) 2 � � = � − (4x + 6x) When (�� + �) = 0, � = − . − 4x + 9 Now, since −(4x − 6) �(�) = �, when � = − , we conclude that 15 � � − = � � We can deduce that: We are now in a position to restate the remainder theorem when the divisor is of the form ax+ b . 2� + 7� + 2� + 9 = (� + 2� − 2)(2� + 3) + 15 ()

Consider (2� + 3) = 0, The Remainder Theorem �(�) = (� + 2� − 2) × 0 + 15 = 15 If fx() is any polynomial and fx() is divided by But, (2� + 3) = 0 implies that � = − . æöb ax+ b then the remainder is f - . Therefore, when � = − , �(�) = 15 or () ç÷ èøa � − = 15. æöb If f ç÷- = 0, then()ax+ b is a factor of fx() . We can conclude that when the polynomial èøa 2� + 7� + 2� + 9 is divided by (2� + 3), the remainder is � − = 15 We apply the Remainder Theorem to obtain the Now consider the example below. remainder when �(�) = 2� + 7� + 2� + 9 was divided by (2� + 3). 2 By the Remainder Theorem, the remainder is 3x + 6x −1 3 2 � − . 3x −1 9x +15x − 9x +1 3 3 3 3 3 2 � − = 2(− ) + 7(− ) + 2(− ) + 9 − (9x − 3x ) 2 2 2 2 2 3 27 63 18x − 9x +1 � − = − + − 3 + 9 2 2 4 4 − (18x − 6x) 3 3x 1 � − = 15 − + 2 − (−3x +1) We can apply the Factor Theorem to show that 0 (3� − 1) is a factor of �(�), where �(�) = 9� + 15� − 9� + 1 We can deduce that: By the Factor Theorem (3� − 1) is a factor of �(�). if � = 0, 9� + 15� − 9� + 1 = (3� + 6� − 1)(3� − 1) �(�) = 9� + 15� − 9� + 1 ( ) 1 1 1 1 Consider 3� − 1 = 0, � = 9 + 15 − 9 + 1 �(�) = (3� + 6� − 1) × 0 = 0 3 3 3 3 But, (3� − 1) = 0 implies that � = . www.faspassmaths.com 1 1 5 Therefore, when � = , �(�) = 0 or � = 0. � = + − 3 + 1 = 0 3 3 3 We can conclude that when the polynomial ( 9� + 15� − 9� + 1 is divided by (3� − 1), the Hence, 3� − 1) is a factor of �(�). remainder is � = 0. Instead of performing long division, we can apply the So, (3� − 1) is a factor of 9� + 15� − 9� + 1 remainder theorem to find the remainder when a polynomial is divided by a linear expression of the From the above examples, we can formulate the form (ax + b). The remainder, � = � − when the following expression: polynomial is divided by the linear factor. �(�) = �(�) × (�� + �) + �

Copyright © 2019. Some Rights Reserved. www.faspassmaths.com Pg 4 of 7 3x2 + 6x + 8 We can use the factor theorem to show that a linear 3 2 expression of the form (ax + b) is a factor of a 2x − 4 6x + 0x − 8x + 5 polynomial. In this case, we show that the remainder − (6x3 −12x2 ) � = � − is zero when the polynomial is divided 2 12x − 8x + 5 by the linear factor. − (12x2 − 24x) 16x + 5 −(16x − 32) Example 3 37 Find the remainder when 32 fx() =+4231 x x- x+ is divided by ()21x - . The quotient is 3� + 6� + 8 Solution The Remainder is 37 In this example ()21x - is of the form ()ax+ b , Factorising and Solving Polynomials where a = 2 and b = -1 We can use the factor theorem to factorise or solve a The remainder would be polynomial. However, to factorise a polynomial of the æö--()1 æö1 form ax32+++ bx cx d it would be helpful to know ffç÷= ç÷. èø22èø one linear factor. Then we can obtain the other factors 32 by the process of long division. æö11 æö æö 1 æö 1 f ç÷=+4231 ç÷ ç÷- ç÷+ èø22 èø èø 2 èø 2 Example 5 11 1 1 Show that (2x + 3) is a factor of =+-11+= 22 2 2 fx() =+2323 x32 x-- x .

Alternatively, we could have used long division to show that the remainder is one half. Solution When f(x) is divided by ()23x + , the remainder is 2 1 3332 3 3 2xx+ 2 -2 f ()2()3()2()30-22=- +- 2 --- 2= 2142xxxx-32+-- 31 Hence, ()23x + is a factor of fx() since the

32 remainder is 0. --42xx 43xx2 - Example 6 Factorise, �(�) = 2xx32+ 3-- 29 x 60 and hence --42xx2 solve fx() = 0 . -x +1

1 --x +2 Solution Let fx()=+ 2 x32 3 x-- 29 x 60. 1 2 To obtain the first factor, we use the remainder theorem to test for f(1), f(-1) and so on, until we Example 4 obtain a remainder of zero. State the quotientwww.faspassmaths.com and the remainder when 6� − 8� + 5 is divided by by 2� − 4 We found that, f (4)=+ 2(4)32 3(4)-- 29(4) 60=+ 128 48-- 116 60= 0

�(4) = 0 Solution Therefore (x – 4 ) is a factor of f(x). In this example, the dividend has no terms in �. It is advisable to add on such a term to maintain Now that we have found a first factor, we divide consistency between the quotient and the divisor. This f(x) by ( x – 4). is done by inserting a term in �as shown below.

Copyright © 2019. Some Rights Reserved. www.faspassmaths.com Pg 5 of 7 2xx2 ++ 11 15 Example 8 xxxx-4232+ 3-- 29 60 If ()x + 3 and ()x - 4 are both factors of 32 -- 2xx32 8 232960xx+-- x , find the third factor.

2 11xx- 29 Solution --11xx2 44 Let the third factor be (�� + �). 15x - 60 We can write the expression as a product of three linear factors as shown. --15x 60

0 2329603xx32+-- x=+()()() x x- 4 axb+ 2 21115325xx++=+()() x x + xxaxx´´ =2 3 \fx()()()()=++25 x x 3 x- 4 \a =2 If fx() = 0, then ()()()25xxx++ 3- 40=. And 34´- ´b =- 60 So � = −3, 2 and 4. \b =5

Example 7 Hence, (ax+ b ), the third factor, is ()25x + . Factorise xx32-+4 x- 12 and hence show that ()()xx+34-= xx2 -- 12 32 xx-+4 x- 12= 0 has only one solution. Alternatively, we may divide 2xx32+ 3-- 29 x 60 by xx2 --12 . Solution To get the first factor, we apply the remainder 25x + theorem as follows: xx232--12 2 x+ 3 x-- 29 x 60 f ()()()()1= 132- 1+ 4 1-¹ 12 0 ---2224xx32 x 32 f -1=- 1 -- 1+ 4-- 1 12 ¹ 0 2 ()()()() 5560xx-- 32 f ()()()()2= 2- 2+ 4 2- 12= 0 ---5560xx2 Hence, (2)x - is a factor of fx() . 0 Dividing by the factor (2)x - \The third factor is ()25x + . xx2 ++6 xxxx--232+ 4- 12 Example 9 32 --xx322 Solve for x in xxx--2 7+= 12 0 , giving the answer to 2 decimal places where necessary. xx2 +4- 12

--xx2 2 Solution 6x - 12 Recall: If fx() is any polynomial and fx() is --6x 12 divided by xa- , then the remainder is fa. 0 () () 32 If the remainder fa() = 0 , then ()xa- is a factor xx-+412 x-www.faspassmaths.com =(2)(xxx- 2 ++ 6) of fx() .

2 First test to see if is a factor. Þ-(2)0xxx= or ( ++= 6)0 (� − 1)

Þx =2 32 Let fx() = x--2712 x x+ The quadratic xx2 ++6 has no real roots because 32 the discriminant f ()()()()1= 1-- 2 1 7 1+ 12¹ 0 22 bac-4(1)4(1)(6)125240=- =-=-< \-()x 1 is not a factor of fx() . 32 Hence, xx-+4 x- 12= 0 has only one solution, x = 2 Next, test to see if (� + 1) is a factor.

f ()()()()-1=----- 132 2 1 7 1+ 12¹ 0 \()x +1 is not a factor of fx() .

Now, try (� − 2) f ()()()()2= 232-- 2 2 7 2+ 12¹ 0 \-()x 2 is not a factor of fx() .

Try (� + 2) f ()()()()-2=----- 232 2 2 7 2+ 12¹ 0 \()x +2 is not a factor of fx() .

Try (� − 3) f ( 3) = (3)3 – 2(3)2 -7(3) +12 = 0 \-()x 3 is a factor of fx() .

Now we divide fx() by ()x -3 . xx2 +-4 xxxx---3271232 + --xx323 xx2 -712+ --xx2 3 -412x + --412x + 0

Since we were asked to give our answers correct to 2 decimal places, we deduce that xx2 +-4 = 0 would not have exact roots. We, therefore, employ the quadratic equation formula to find the roots.

2 2 -bb± -4 ac If ax++= bx c 0 , then x = , 2a where � = 1, � = 1 and � = −4.

www.faspassmaths.com2 -()()()()11414± -- x = 21() -117± = 2 =1.561 or- 2.561 =1.56 or- 2.56 (correct to 2 decimal places)

Hence, x =3, 1.56 or- 2.56 .