Just put the final answer on the box;

solution) 4 2. 2 x4 + x 3 - 16x2 +18 - r + 2 3. Lesson 4: Factor Theorem and its Converse Definition: The binomial x - c is a factor of a polynomial p(x) if the remainder is 0 when p(x) is divided by x - c. Conversely, if the remainder when p(x) is divided by x - c is 0, then x - c is factor of p(x). 8Illustrative Example: Determine which of the following binomials are factors of f(x) = 2x3-7x2-5x+4. a. x + 1 b . x - 2 c. x - - Solution: a. Equate the binomial to O to find c. x+1=0 x =-1 f (x) = 2x3-7x2 -5x+4 f ( - 1) = 2(-1)3 - 7(-1)2 -5(-1) +4 = 2(-1) -7(1) -5(-1) +4 =-2-7+5+4 = 0 Hence, the remainder is 0. By Factor Theorem, since f(-1) = 0, therefore x + 1 is a factor of f (x). b. Equate the binomial to O to find c. x - 2= 0 x = 2 f(x) = 2x3-7x2 -5x+4 f ( 2 ) = 2(2) 3 - 7 ( 2 ) 2 - 5 ( 2 ) + 4 = 2(8) -7(4) -5(2) +4 =16 -28 - 10+4 -18 Hence, the remainder is -18. By Factor Theorem, since f (2) = -18, therefore x - 2 is a not factor of f (x ). c. Equate the binomial to 0 to find c. -1 = 0 f ( xx ) = 2(#)3- 7 ( 4) - 5(#)+ 4 f () = 2()-7()-5()+4 - 2+4 1- 7 - 10 + 16 = 0 Hence, the remainder is 0. By Factor Theorem, since f () = 0, therefore x - } is a factor of f(x). Activity 5. Determine if x - 1 and x + 1 are factors of each polynomial. 1. f (x ) = x2-x+2 3. f(x) =-3x3-3x 2. f (x ) =-x3+1 4. f(x) =-x2+2x-1 Bonus Question: Find m so that dividing x4 - mx3 - 2x2 + (3m - 5)x + m by (x - m) gives a remainder of 5.\fActivity 5 1 ) P ( x ) = X 2 - X + 2 ) f - + 0 - ex m - X - 1 = 0 x + 1 = 0 = ( 1 ) = ( 1 ) 2 - 1 +2 wolonion F. ( - 1 ) = ( - 1) 2 - 1+ 2 P (1 ) = 1 - 1 + 2 F ( - 1 ) = 1 - 1 + 2 F ( 1 ) = 2 F C - 1) = 1 - 2 1 The remainder is 2: to sulior The tremainder is 2. * ( 2 - me ) + +x8 - sam - by By (Factor * Theorem ) since of ( 1) = (12) and x p (- 1)= 2, there Fore, * = 1 - and x t/+ cire " not factor OF F ( x ) = x 2 - x + 2 . 1 - +8- 5 x8 - dxt. + x 2 . ) XP ( x ) X= - x3+ +x/ nortonNg . wow svart sul X - 1 = 0 * +1 = 0 F ( 1 ) = - ( 1)3 + 1 0 = M - F ( - 1 ) = - (- 173 + 1 F ( 1 ) = - 1 + 1 O = ( 1 - ) -F X ( - 1 ) = - ( - 1) + 1 P ( 1) . = 0 0 = 1+ + ( - 1 ) - 1 + 1 The remainder is O. 1 - = * + ( - 1 ) = 2 1 - x8 - XS - Theremainder is 2 . By Factor theorem , since St (i) = 0 and F ( - 1) = 2, therefore, x - It is a Factor OF + ( x ) and x + 1 is not a factor of $ ( x ) . of3 . ) + ( * ) = - 3 * 3 - 3 x X - 1 = 0 *+1 =0 X = 1 * = - 1 $ ( 1 ) = 3( 1 ) 3 - 3 ( 1 ) F ( - 1 ) = - 3 ( - 1)3 - 3 ( - 1 ) # ( 1 ) = - 3 (1) - 3 F ( - 1 ) = - 3 ( - 1) + 3 # ( 1 ) = - 3 - 3 F ( - 1 ) = 3 + 3 # ( 1 ) = - 6 P ( - 1) = 6 The remainder is -6. The remainder is 6. By Factor theorem, since , ( 1) = -6 and $ (-1)= 6, therefore, X-1 and x +1 are not factor OF + ( x ) = - 3 x 3 - 3x. 4.) + ( * ) = - x 2 + 2x - 1 X - 1 = 0 x+ 1 20 X = 1 x = - ) * ( 1 ) = - ( 172 + 2 ( 1 ) - 1 F ( - 1 ) = - (- 1)2 + 2(-17 - 1 + ( 1 ) = - 1 +2 - 1 * ( - 1 ) = - (1)-2 - 1 + ( 1 ) = - 2 + 2 I (- 17 - - 1 - 2- 1 + ( 1 7 = 0 FC - 17 = - 4 The remainder is . The remainder is - 4. By Jactor theorem , vince + ( 1 ) = 0 and # (- 1 ) = - 4, therefore , x - 1 is a factor of I (x ) and x+ 1 is not a factor of tex )