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Mathematics 10 Online
OpenStudy (embracetheweebness98):

What polynomial has roots of −6, −4, and 1?

OpenStudy (lord_box):

You can build a polynomial with factors. For instance, a polynomial with roots 2 and 3: x = 2, 3 (x - 2)(x-3) = 0 x^2-5x+6 = 0

OpenStudy (embracetheweebness98):

oh okay

OpenStudy (embracetheweebness98):

how does that work? like, the roots and stuff?

OpenStudy (welshfella):

a factor when root is 3 will be (x - 3) Because when x - 3 = 0 x = 3 if root is -2 the factor is ( x + 2) Because when x + 2 = 0 x = -2

OpenStudy (welshfella):

so a polynomial with roots -2 and 3 is (x + 2)(x - 2)

OpenStudy (welshfella):

So do same process with -6,-4 and 1. You will have 3 factors

OpenStudy (embracetheweebness98):

oh okay, thank you!

OpenStudy (embracetheweebness98):

it makes so much sense now XD

OpenStudy (welshfella):

yw

OpenStudy (embracetheweebness98):

I have (x+6) (x+4) (x-1)

OpenStudy (lord_box):

that's right, you have to expand it for it to be a polynomial, though.

OpenStudy (embracetheweebness98):

okay

OpenStudy (embracetheweebness98):

I foil?

OpenStudy (embracetheweebness98):

@Lord_Box

OpenStudy (lord_box):

Yes

OpenStudy (phi):

First do (x+6) (x+4) x^2 + 4x+6x+24 or x^2 +10x+24 now multiply that by (x-1) one way is to distribute (x-1) for example (x-1)(x^2 +10x+24) means multiply *each* term inside the parens by (x-1) (x-1)x^2 + (x-1)*10x + (x-1)*24 now simplify each term (distribute again!) and collect terms can you do that ?

OpenStudy (embracetheweebness98):

ok

OpenStudy (embracetheweebness98):

(x^3-1x^2) + (10x^2-10x) + (24x-24) ?

OpenStudy (lord_box):

That's right, and combine your like terms.

OpenStudy (embracetheweebness98):

I got x^3+9x^2+14x-24

OpenStudy (phi):

looks good

OpenStudy (phi):

now you can try to "go backwards" and try to figure out the roots of that equation i.e. what x values make it zero. That is a much harder problem (but one you will often do for quadratic i.e. power of 2 (not 3 like here))

OpenStudy (embracetheweebness98):

okay

OpenStudy (embracetheweebness98):

so i start like this? x^3+9x^2+14x-24= 0

OpenStudy (phi):

yes, but it's a hard problem (I don't think people even teach how to do it by hand) Don't bother unless you are a math nerd.

OpenStudy (embracetheweebness98):

What should i do now then? I still have to figure out the answer

OpenStudy (phi):

the question is **What polynomial has roots of −6, −4, and 1?** the answer is what you found x^3+9x^2+14x-24

OpenStudy (lord_box):

Terminology wise: x^2 is a monomial x^2 + x is a binomial and a polynomial x^2 + x + 1 is a trinomial and a polynomial. polynomials have multiple terms

OpenStudy (embracetheweebness98):

oh, hold up

OpenStudy (embracetheweebness98):

the answer choices are 3x2 + 16x + 42 3x2 + 4x + 2 3x2 − 16x + 42 3x2 − 4x + 2

OpenStudy (phi):

those are the choices to what question ?

OpenStudy (embracetheweebness98):

The original one *What polynomial has roots of −6, −4, and 1?*

OpenStudy (lord_box):

A quadratic equation can have at most 2 roots

OpenStudy (phi):

First, if a polynomial has 3 roots, it has "order 3" i.e. the highest exponent must be 3 (as in we *need* an \( x^3\) term Second, if we sub in x=1 into all 4 choices, none give 0 So we have a problem here. Mention it to your teacher.

OpenStudy (embracetheweebness98):

okay, this is pretty confusing. I'll guess on this one for now, thanks for the help!

OpenStudy (phi):

You probably have a typo or possibly you are misreading the question. Can you post a screen shot ?

OpenStudy (embracetheweebness98):

sure, hold up

OpenStudy (embracetheweebness98):

OpenStudy (phi):

ok the question is **What polynomial has roots of −6, −4, and 1?** the answer is x^3+9x^2+14x-24

OpenStudy (phi):

one of the 4 choices is the answer.

OpenStudy (embracetheweebness98):

oh wait...

OpenStudy (embracetheweebness98):

i must be really tired XD

OpenStudy (embracetheweebness98):

Thank you!

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