• Apr 22nd 2009, 02:04 PM
Maccabhoy
Hi, im doing a Maths quiz, and am only stuck now on a couple more questions where i was told my anwer is wrong....if anyone could show me the light, it would be appreciated.

1) Given:

x^3+1 = ( x+a ) ( x^2+bx+c )

Find the values of a, b and c. Please enter your answer as a list [in brackets], in the form: [a, b, c]

2) Factorise the following polynomial completely over the integers.

3 x^4+18 x^2+15
• Apr 22nd 2009, 02:33 PM
Prove It
Quote:

Originally Posted by Maccabhoy
Hi, im doing a Maths quiz, and am only stuck now on a couple more questions where i was told my anwer is wrong....if anyone could show me the light, it would be appreciated.

1) Given:

x^3+1 = ( x+a ) ( x^2+bx+c )

Find the values of a, b and c. Please enter your answer as a list [in brackets], in the form: [a, b, c]

2) Factorise the following polynomial completely over the integers.

3 x^4+18 x^2+15

1. Use the factor theorem.

If $\displaystyle P(x) = x^3 + 1$ and $\displaystyle (x + a)$ is a factor, then $\displaystyle P(-a) = 0$.

So $\displaystyle (-a)^3 + 1 = 0$

$\displaystyle -a^3 + 1 = 0$

$\displaystyle 1 = a^3$

$\displaystyle a = 1$ (I assume you're factorising over the reals...)

So we have $\displaystyle x^3 + 1 = (x + 1)(x^2 + bx + c)$.

Now use long division of $\displaystyle \frac{x^3 + 1}{x + 1}$ or the sum of two cubes rule to find what b and c are.
• Apr 22nd 2009, 02:35 PM
Prove It
Quote:

Originally Posted by Maccabhoy
Hi, im doing a Maths quiz, and am only stuck now on a couple more questions where i was told my anwer is wrong....if anyone could show me the light, it would be appreciated.

1) Given:

x^3+1 = ( x+a ) ( x^2+bx+c )

Find the values of a, b and c. Please enter your answer as a list [in brackets], in the form: [a, b, c]

2) Factorise the following polynomial completely over the integers.

3 x^4+18 x^2+15

$\displaystyle 3x^4 + 18x^2 + 15$

This is a quadratic if written with $\displaystyle X = x^2$

$\displaystyle = 3X^2 + 18X + 15$

$\displaystyle = 3(X^2 + 6X + 5)$

$\displaystyle = 3(X + 5)(X + 1)$

$\displaystyle = 3(x^2 + 5)(x^2 + 1)$

These can now only be factorised over Reals, so this is as far as you need to go.
• Apr 22nd 2009, 02:54 PM
Maccabhoy
Quote:

Originally Posted by Prove It
1. Use the factor theorem.

If $\displaystyle P(x) = x^3 + 1$ and $\displaystyle (x + a)$ is a factor, then $\displaystyle P(-a) = 0$.

So $\displaystyle (-a)^3 + 1 = 0$

$\displaystyle -a^3 + 1 = 0$

$\displaystyle 1 = a^3$

$\displaystyle a = 1$ (I assume you're factorising over the reals...)

So we have $\displaystyle x^3 + 1 = (x + 1)(x^2 + bx + c)$.

Now use long division of $\displaystyle \frac{x^3 + 1}{x + 1}$ or the sum of two cubes rule to find what b and c are.

Thanks for the info. Much appreciated. But sorry for being a bit slow, but not sure what to divide to get b and c
• Apr 22nd 2009, 03:26 PM
Maccabhoy
Could I also get a hint for this....im actually not just sure, if its my syntax for the quiz program giving me the wrong answer...

Find the constants a and b such that
x^2+3 x+a = ( x+3 ) ( x+b )
• Apr 22nd 2009, 05:09 PM
Prove It
Quote:

Originally Posted by Maccabhoy
Could I also get a hint for this....im actually not just sure, if its my syntax for the quiz program giving me the wrong answer...

Find the constants a and b such that
x^2+3 x+a = ( x+3 ) ( x+b )

Expand the right hand side.

You should get $\displaystyle x^2 + 3x + a = x^2 + (3 + b)x + 3b$

You should be able to find $\displaystyle a$ and $\displaystyle b$ now.

Also for the question $\displaystyle x^3 + 1 = (x + a)(x^2 + bx + c)$, it might be easiest to use the sum of two squares rule.

$\displaystyle \alpha ^3 + \beta ^3 = (\alpha + \beta)(\alpha ^2 - \alpha \beta + \beta^2)$

In your case, $\displaystyle \alpha = x, \beta = 1$.

Have a go.