A-Math Remainder Theorem

Jt061952 SG

I am trying to work on this problem:

Find the value of a and b for which x^2 + 4x -1 is a factor of x^3 + ax^2 - 5x + b

The roots of x^2 + 4x -1 are 0.236.... and 4.236... Because of these decimal numbers, I cannot get a and b as whole numbers. a = 3, b = 1. Is there a better way? Thanks. jt

Forbiddensinner SG

I am trying to work on this problem:

Find the value of a and b for which x^2 + 4x -1 is a factor of x^3 + ax^2 - 5x + b

The roots of x^2 + 4x -1 are 0.236.... and 4.236... Because of these decimal numbers, I cannot get a and b as whole numbers. a = 3, b = 1. Is there a better way? Thanks. jt

Divide ( x^3 + ax^2 - 5x + b ) by ( x^2 +4x - 1 ) using long division method.

You should get quotient ( x + a - 4 ) and a negligible remainder, as it won't help you in solving the problem.

Next, you multiply the quotient ( x + a - 4 ) with the divisor ( x^2 + 4x -1 ), and you should get ( x^3 + ax^2 + (4a-17)x + 4 -a ).

Finally, use the new cubic equation you have and equal it to the original cubic equation, and after simplifying the numbers on both side ( NOTE : DO NOT BRING ANYTHING OVER FROM EITHER SIDE ), you should get:

-5x + b = (4a - 17)x + (4-a)

Now, you can directly compare coefficients to get

-5 = 4a - 17 and b = 4 - a

Hence, a=3 and b=1.

nightzip SG

If x^{2 + 4x -1 is a factor of x}3 + ax^{2 – 5x +
b,}

then finding roots for x2 + 4x ~~1, using discriminant formula, x=-2 +/~~ sqrt (5)

Let f(-2+sqrt(5))= (-2+sqrt(5))^{3 + a(-2+sqrt(5))}2 – 5(-2+sqrt(5)) + b = 0

Similarly, for f(-2-sqrt(5)) = (-2-sqrt(5))^{3 +
a(-2-sqrt(5))}2 – 5(-2-sqrt(5)) + b = 0

Since two equations, two unknowns, you can solve for a and b.

Jt061952 SG

Thanks, that was quick. Now I know how to do both methods. Appreciate it.

wee_ws SG

Hi,

There is an even faster approach:

(x^2 + 4x - 1)(x - b) = x^3 + ax^2 - 5x + b.

Comparing coefficients of x^2, we obtain -b + 4 = a -- (1).

Comparing coefficients of x, we obtain -4b - 1 = -5 -- (2).

Now we can solve for a and b.

Thanks!

Cheers,
Wen Shih

Forbiddensinner SG

Hi,

There is an even faster approach:

(x^2 + 4x - 1)(x - b) = x^3 + ax^2 - 5x + b.

Comparing coefficients of x^2, we obtain -b + 4 = a -- (1).

Comparing coefficients of x, we obtain -4b - 1 = -5 -- (2).

Now we can solve for a and b.

Thanks!

Cheers,
Wen Shih

Nice, as expected from a homework forum mod.

To the TS, the reason why this method can be used here is because the original cubic equation did not contain any other numbers besides 'b', hence wee made use of this to determine (x-b) as one of the roots.

Jt061952 SG

Thanks Wen Shih, your method is more elgant. I find substituting sqrt to be very tedious. For a 5 marks problem, it should not take that long.

nightzip SG

Wow Wen Shih, that a very good solution.
However, didnt quite understand how (x-b) can be so quickly seen as a root?

Lee012lee SG

I am trying to work on this problem:

Find the value of a and b for which x^2 + 4x -1 is a factor of x^3 + ax^2 - 5x + b

The roots of x^2 + 4x -1 are 0.236.... and 4.236... Because of these decimal numbers, I cannot get a and b as whole numbers. a = 3, b = 1. Is there a better way? Thanks. jt

Purepls aka Jt061952 huh ?

Same question asked twice.

Once, the question is asked in the thread "AMaths" by Purepls

Again, the same question in this thread "AMaths Remainder Theorem" by
Jt061952 .

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(1) Find the value of a and b for which x^2 - 2x - 3 is a factor of

ax^3 - 8x^2 +bx + 6

(2) Find the value of a and b for which x^2 + 4x -1 is a factor of

x^3 + ax^2 - 5x + b

(3) Find the value of a and b for which x^2 + x + 4 is a factor of

x^3 + ax^2 - 5x + b

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All three questions look alike.

Using Nightzip's method, it will be lengthy to solve question 2 and the answers of a and b will not be exact too. While nightzip's method will not be able to solve question 3.

Mr Wee and Mikethm method can solve question 3, 2 and 1.

wee_ws SG

Wow Wen Shih, that a very good solution.
However, didnt quite understand how (x-b) can be so quickly seen as a root?

Hi,

Thanks for your encouragement :)

(x - b) has to be a factor because the expression has b as a constant.

Thanks again.

Cheers,
Wen Shih