1. ## Factorising quadratic polynomials

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

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

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

$-a^3 + 1 = 0$

$1 = a^3$

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

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

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

3. 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
$3x^4 + 18x^2 + 15$

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

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

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

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

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

These can now only be factorised over Reals, so this is as far as you need to go.

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

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

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

$-a^3 + 1 = 0$

$1 = a^3$

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

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

Now use long division of $\frac{x^3 + 1}{x + 1}$ 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

5. 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 )

6. 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 $x^2 + 3x + a = x^2 + (3 + b)x + 3b$

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

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

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

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

Have a go.