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Class Number 13 – 16 IV – St. Hannibal 1. Find m if x3 + 3x2 – mx + 4 leaves a remainder of m when divided by x – 2.
23 + 3(2)2 – m(2) + 4 = m

8 + 12 – 2m +4 = m

-3m = -24

-3 -3 m = 8

2. When the expression 6x2 – 2x +3 is divided by x –a, the remainder is 3. Find the value of a.
6a2 – 2a + 3 = 3

6a2 – 2a = 0

6a2 = 2a
6a 6a

a =

3. The expression x4 + x3 + 2mx2 – 14m4 leaves a remainder of 32 when divided by x + 2m. Find the possible values of m.

(-2m)4 + (-2m)3 + 2m(-2m) – 14m4 = 32

16m4 – 8m3 + 8m3 – 14m4 = 32

2m4 = 32

2 2
4 =

m = 2

4. When the expression 8x3 + mx2 + nx – 9 is divided by x + 2 and 2x – 3 the remainders are -93 and 6, respectively. Find the value of m and n.
8x3 + mx2 + nx – 9 = -93 8x3 + mx2 + nx – 9 = 6

8(-2)3 + m(-2)2 + n(-2) – 9 = -93 8()3 + m()2 + n() – 9 = 6

-64 + 4m – 2n – 9 = -93 18 + + = 6

4m – 2n = -20 + = -12

(4m – 2n = -20) 3 4m – 2n = -20
( + = -12) 4 4() – 2n = -20
12m – 6n = -60 – 2n = -20
9m + 6n = -48 -2n = -20 + 21m = -108 (-2n = ) 21 21 m = n= -2/7 5. The polynomial x3 + mx2 +nx – 3 leaves a remainder of 21 when divided by x – 2 and a remainder of 3 when divided by x + 1. Calculate the remainder when the polynomial is divided by x – 1. x3 + mx2 + nx – 3 = 21 x3 + mx2 + nx – 3 =3

23 + (m2)2 + n(2) – 3 = 21 (-1)3 + m(-1)2 + n(-1) -3 = 3

4m + 2n = 16 -1 + m – n -3 = 3

m – n + 7

4m + 2n = 16 4m + 2n = 16 m – n = 7 R = (1)3 + 5(1)2 + (-2)(1) - 3

(m – n = 7) 2 2m – 2n = 14 n = -2 =1 + 5 -2 – 3 6m = 30 R = 1

6 6 m = 5

6. The expression x3 + ax2 +7 leaves a remainder p – 1 when divided by x + 1 and a remainder of p + 2 when divided by x – 2. Find the value of a and p.
(-1)3 + a(-1)2 + 7 = p – 1 (2)3 + a(2)2 + 7 = p + 2 (4a – p = -13)-1

-1 + a +7 = p – 1 8 + 4a + 7 = p + 2 a – p = -7

a – p = -7 4a – p