Remainder Theorem and possibly Simultaneous Equations?

if

6x^3+ax^2+bx+ 4, when divided by x+ 1 the remainder is -15

then

p(x) = ( 6x^3+ax^2+bx+ 4) - (-15) is dividable by (x+1)

so

the division gives :

p(x)= (6x^2 + (a-b)x - (a-b-6)) (x + 1)

with the remainder :

a - b + 13 = 0 -------> equation (1)

as

p(x) = ( 6x^3+ax^2+bx+ 4) - 49 is dividable by (x-3)

so

the division gives :

p(x)= (6x^2 + (a+18)x + (3a+b+54) ) (x - 3)

with the remainder :

3(3a + b + 54) - 45 = 0

9a + 3b + 117 = 0

3a + b + 39 = 0 -------> equation (2)

replacing by its value b = a + 13

in eq(1) gives :

3a + a + 13 + 39 = 0

4a = - 52

a = -13 and b = 13 - 13 = 0

et voilà !!

Mad's solution is "lighter" !! ;-)

hope it' ll help !!

When 6x^3 + ax^2 + bx + 4 is divided by x + 1 the remainder is --15 whence

6(--1)^3 + a(--1)^2 + b(--1) + 4 = -- 15 OR a -- b = -- 13 ..........(A)

and when the expression is divided by x −3 the remainder is 49 whence

6(3^3) + a(3^2) + b(3) + 4 = 49 OR 3a + b = -- 39 .............(B)

Adding (A) and (B), we get 4a = -- 52 OR a = -- 13 ANSWER

When a = -- 13, from (B), b = -- 39 -- 3a = -- 39 -- 3(--13) = 0 ANSWER

f(x) = 6x^3 + ax^2 + bx + 4

f(- 1) = - 6 + a - b + 4 = a - b - 2 = - 15

f(3) = 162 + 9a + 3b + 4 = 9a + 3b + 166 = 49

equation 1: a - b = - 13

equation 2: 9a + 3b = - 117

I would solve with substitution: a = (b - 13)

You can go with elimination if you want !! You can do this now !!

MAD

[a = - 13, b = 0 ]