Factoring a Polynomial into Linear and Quadratic Factors: P(x)=x^2+8x^2-9

Illvoices

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(A) Factor P into linear and irreducible quadratic factors with real coefficients. (b) factor P completely into linear factors complex coefficient.

P(x)=x^2+8x^2-9
 

ksdhart2

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You've posted a bunch of problems (I honestly don't even know how many; I've lost count) and most of the time you don't post any work with them, until prodded to do so. This tells me you didn't read the rules of the forum as outlined in the Read Before Postinghttps://www.freemathhelp.com/forum/threads/54004-Read-Before-Posting thread that's stickied at the top of every sub-forum. You need to do so now. After reading the rules, please comply with them by sharing with us any and all work you've done on this problem, including the parts you know for sure are wrong. Then we'll help you figure out how best to proceed. Thank you.
 

HallsofIvy

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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 "\(\displaystyle 8x\)".)
 
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Illvoices

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The polynomial is actually:
P(x)=x2+8x2-9

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 :?
 

MarkFL

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The polynomial is actually:
P(x)=x2+8x2-9

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

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

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. :D
 

HallsofIvy

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The polynomial is actually:
P(x)=x2+8x2-9

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

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 :?
 

MarkFL

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I think it's likely that the problem given to the OP has a typo in it, and given the title of the thread (a cubic polynomial could have a linear AND a quadratic factor), I would posit one of these was the intended problem:

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. :D
 
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