Question

F(x)=x^3+ax^2+bx+3 where a and b are constants. bx Given that when f (x) is divided by (x+2) the remainder is 7, (a) show that 2a-b=6 Given also that when f (x) is divided by (x-1) the remainder is 4. (b) Find the value of a and the value of b

I don't know how solve it :(((((((

Answer 1

Part I - First synthetic division

You need to use synthetic division to come up with an expression for a and b:

(x + 2) is a factor, and the remainder is 7, so we can draw a synthetic division table...

coefficients = 1 for X^3; A for X^2; B for X^1; and 3

-2 | 1 A B 3

-2 -2(A-2) 4(A-2)-2B

1 (A-2) -2(A-2)+B 4(A-2)-2B + 3

Remainder = 7

So...

4(A-2)-2B + 3 = 7

4 * (A - 2) - 2B + 3 = 7

4A - 8 - 2B = 4

4A - 2B = 12

2A - B = 6
Proved

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Part II - Second Synthetic Division

We draw another synthetic division table, this time with (x - 1), so the number on the left hand side will be +1

1 | 1 A B 3

1 (A+1) A+B+1

1 (A+1) A+B+1 A+B+4

Remainder = 4

So...

A + B + 4 = 4

A + B = 0

A = -B

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Part III - Solving for A and B with our two simultaneous equations

We know that A = -B and we also know that 2A - B = 6

Since we know that A is equal to -B We can substitute in A for -B, to get:

2A - B = 6

Therefore...

2A + A = 6

3A = 6

A = 2

Again, as we know that A = -B, and as we have found that A = 2, we can see:

A = -B

Therefore...

2 = -B

B = -2

So our final answer is A = 2, B = -2

Hopefully this answer is more useful than the last one, and isn't so confusing!