- Thread starter Illvoices
- Start date

- Joined
- Jan 27, 2012

- Messages
- 5,046

You could start by "completing the square" or using the "quadratic formula" to solve \(\displaystyle x^2+ 8x- 9= 0\). This is also very easy to factor by considering factors of 9.

(I assume your "\(\displaystyle 8x^2\)" was supposed to be "\(\displaystyle 8x\)".)

(I assume your "\(\displaystyle 8x^2\)" was supposed to be "\(\displaystyle 8x\)".)

Last edited:

P(x)=x

so what i'd first do is factor 9 and that was going to give me 3*3, but then i thought how is the factor of nine the product of 8 or in this case how is the factor of eight 4,2 or those this mean that you have to find the relation between these values by adding 4,2,and 8. I'm not so good at factoring btw :?

- Joined
- Nov 24, 2012

- Messages
- 1,828

In this case, you want to combine like terms first:

P(x)=x^{2}+8x^{2}-9

so what i'd first do is factor 9 and that was going to give me 3*3, but then i thought how is the factor of nine the product of 8 or in this case how is the factor of eight 4,2 or those this mean that you have to find the relation between these values by adding 4,2,and 8. I'm not so good at factoring btw :?

\(\displaystyle P(x)=9x^2-9\)

Next, you can factor out 9:

\(\displaystyle P(x)=9(x^2-1)\)

And now you can proceed using the difference of squares to finish.

- Joined
- Jan 27, 2012

- Messages
- 5,046

No, it doesn't give just 3*3. 9 can be factored as 3*3 or 1*9. -9 can be factored as 3*-3 or -1*9 or 1*-9.The polynomial is actually:

P(x)=x^{2}+8x^{2}-9

so what i'd first do is factor 9 and that was going to give me 3*3

Which of those gives 8x? (The factors of 8 are irrelevant. \(\displaystyle (x+ a)(x+ b)= x^2+ (a+b)x+ ab\) so you want ab= -9 and a+b= 8

, but then i thought how is the factor of nine the product of 8 or in this case how is the factor of eight 4,2 or those this mean that you have to find the relation between these values by adding 4,2,and 8. I'm not so good at factoring btw :?

- Joined
- Nov 24, 2012

- Messages
- 1,828

a) \(\displaystyle P(x)=x^3+8x^2-9\)

b) \(\displaystyle P(x)=x^3+8x-9\)

In both cases, we observe that \(\displaystyle P(1)=0\), and so use of either polynomial or synthetic division, will easily yield a factorization.