what are the ration… - QuestionCove
OpenStudy (anonymous):

what are the rational zeros of $6x ^{4}+32x ^{3}-70x ^{2}$

2 years ago
OpenStudy (miracrown):

$f(x) = 6x^4 + 32x^3 - 70x^2$ Let's start by factoring out the GCD, greatest common divisor For example, everything up here is divisible by 2x^2, right?

2 years ago
OpenStudy (anonymous):

yeah. that's what i was thinking of doing first, but now i don't know what to do.

2 years ago
OpenStudy (anonymous):

so i would get an answer of 2x^2(3x^2+16x-35)?

2 years ago
OpenStudy (miracrown):

That would be our first factorization, yes

2 years ago
OpenStudy (anonymous):

Okay. So what would I do next?

2 years ago
OpenStudy (miracrown):

$2x^2(3x^2+16x-35)$ So we at least know it has a zero at x = 0, but we might have rational zeros of that quadratic. Just to make sure: The question asked for all _actual_ rational zeros, right? Not just _possible_ rational zeros?

2 years ago
OpenStudy (anonymous):

The exact question is: Find all the rational zeros of the function...

2 years ago
OpenStudy (miracrown):

K gotcha! So now we know we have a rational zero of x = 0, but we might get more from that quadratic in parenthesis, right?

2 years ago
OpenStudy (anonymous):

Yep.

2 years ago
OpenStudy (miracrown):

$2x^2 = 0$ $x = 0$ $3x^2 + 6x -35 = 0$

2 years ago
OpenStudy (miracrown):

So we can either (attempt to) factor, or use the quadratic formula

2 years ago
OpenStudy (anonymous):

Do you think that doing the quadratic formula is easiest?

2 years ago
OpenStudy (miracrown):

And I just went for computing the discriminant, to see if it could be factored: 6^2 - 4(3)(-35), which wound up being 676, which is indeed 26^2

2 years ago
OpenStudy (miracrown):

16^2, not 6^2, I meant, sorry

2 years ago
OpenStudy (miracrown):

I do, yeah. Otherwise we'd be looking for a pair of numbers that multiply to -105 that add up to 16, or "guess and check", and that all seems harder than just the quadratic formula

2 years ago
OpenStudy (anonymous):

Okay.

2 years ago
OpenStudy (miracrown):

$b^2 - 4ac : 16^2 - 4(3) (-35) =676 = 26^2$

2 years ago
OpenStudy (anonymous):

So for my final answers I would get, x= 0, -1.66, and 7?

2 years ago
OpenStudy (miracrown):

Nope. Rational zeros: 0, -8 and 5/3

2 years ago
OpenStudy (anonymous):

How? I'm confused....

2 years ago
OpenStudy (miracrown):

$x = \frac{ -b \pm \sqrt{b^2 - 4ac} }{ 2a }$ $= \frac{ -16\pm \sqrt{26^2} }{ 2(3) }$ $=\frac{ -16\pm 26 }{ 6 }$ $x = \frac{ -48 }{ 6 } = -8$ $x = \frac{ 10 }{ 6 } = \frac{ 5 }{ 3 }$ $Therefore \space \space rational \space \space zeros : 0, \space -8, \space \frac{ 5 }{3 }$

2 years ago
OpenStudy (anonymous):

Oh!!! I missed the negative on the 16. But I'm still confused about how you got -8. I keep getting -42 instead of -48.

2 years ago
OpenStudy (miracrown):

$\frac{ -48 }{ 6 } = -8$

2 years ago
OpenStudy (anonymous):

Isn't -16-26=-42 though? Or am I just confused.....

2 years ago