Algebraic division help

Question

When ax²+3bx+a+2 is divided by (x+2) the remainder is 6, and when bx³ +x²+abx+3b is divided by (x+3) the remainder is 5. What are the values of a and b?

Answer 1

Algebraic division help

When ax2+3bx+a+2 is divided by (x+2) the remainder is 6, and when bx3 +x2+abx+3b is divided by (x+3) the remainder is 5. What are the values of a and b?

Hello Krugen,

Use synthetic division with the coefficients from the equations in order and the factor given as shown:

-2⌋ a ... 3b ........... a+2

.............-2a .......... -6b+4a

———————————

.......a ... 3b - 2a ... 5a - 6b + 2

remainder: 5a - 6b + 2 = 6

-3⌋ b ... 1 ............ ab ............... 3b

........... -3b ......... -3+9b ......... -3ab + 9 - 27b

———————————————

..... b ... 1 - 3b .... ab - 3 + 9b ... -3ab + 9 - 24b

remainder: -3ab + 9 - 24b = 5

Use the two remainder equations to solve for a & b:

5a - 6b + 2 = 6 *solve for a and substitute in 2nd equation.

a = (4 + 6b)/5 Now substitute:

-3ab + 9 - 24b = 5

-3[4 + 6b)/5]*b + 9 - 24b = 5

(-12b - 18b²)/5 + 9 - 24b = 5 *Solve for b:

-12b - 18b² + 45 - 120b = 25

18b² + 132b - 20 = 0

9b² + 66b - 10 = 0

Use quadratic formula:

b = 0.1485

or

b = -7.482

Substitute both answers for b back into this equation a = (4 + 6b)/5, to solve for a:

a = (4 + 6(0.1485))/5 = 0.9782

a = (4 + 6(-7.482))/5 = -8.178

Solutions: (0.9782, 0.1485) and (-8.178, -7.482)