Synthetic division (Finding all zeros)

**shiznid12** · October 14th 2009, 12:37 PM

Show that the given value of c is a zero of P(x), and find all other zeros of P(x) (Enter NONE in any unused answer blanks.)

P(x) = -2x3 + 2x2 + 4x - 4, c = 1.

I figured that all out, and 1 does bring a remainder of 0.

And I believe my end equation is (x-1) (-2x^2+4)

From there, I don't know how to find the other zeros... and my book doesn't help at all.

Thanks guy

**masters** · October 14th 2009, 12:59 PM

Originally Posted by shiznid12

Show that the given value of c is a zero of P(x), and find all other zeros of P(x) (Enter NONE in any unused answer blanks.)

P(x) = -2x3 + 2x2 + 4x - 4, c = 1.

I figured that all out, and 1 does bring a remainder of 0.

And I believe my end equation is (x-1) (-2x^2+4)

From there, I don't know how to find the other zeros... and my book doesn't help at all.

Thanks guy

Hi shiznid12,

$p(x) = -2x^3 + 2x^2 + 4x - 4$

Your depressed polynomial is correct.

$p(x)=(-2x^2+4)(x-1)$

Factor a -2 from the first term.

$p(x)=-2(x^2-2)(x-1)$

Set p(x) =0. Evaluate $x^2-2=0$ to find your other two zeros.

**e^(i*pi)** · October 14th 2009, 01:00 PM

Originally Posted by shiznid12

Show that the given value of c is a zero of P(x), and find all other zeros of P(x) (Enter NONE in any unused answer blanks.)

P(x) = -2x3 + 2x2 + 4x - 4, c = 1.

I figured that all out, and 1 does bring a remainder of 0.

And I believe my end equation is (x-1) (-2x^2+4)

From there, I don't know how to find the other zeros... and my book doesn't help at all.

Thanks guy

Looks right to me. Personally I would factor out a -2 (like masters did) and use the difference of two squares:

$P(x) = -2(x-1)(x-\sqrt{2})(x+\sqrt{2})$

**shiznid12** · October 14th 2009, 01:05 PM

Wow... it's always the tiny steps that ruin everything. I didn't even think to factor out the 2

Thanks a lot guys.

**e^(i*pi)** · October 14th 2009, 01:07 PM

Originally Posted by shiznid12

Wow... it's always the tiny steps that ruin everything. I didn't even think to factor out the 2

Thanks a lot guys.

You don't have to factor the 2, it would have came out when solving $4-2x^2=0$ , it just helps xD

I'm not sure if you know but for $ab=0$ either a and/or b must be 0

**masters** · October 14th 2009, 01:10 PM

Originally Posted by shiznid12

Wow... it's always the tiny steps that ruin everything. I didn't even think to factor out the 2

Thanks a lot guys.

You really don't have to factor out the -2 since all you're interested in are the remaining zeros.

$-2x^2+4=0$

$-2x^2=-4$

$x^2=2$

$x=\pm \sqrt{2}$

Sorry to be redundant, but posted a tad too late. Yeah, what he said!

**shiznid12** · October 14th 2009, 01:10 PM

Well it helped.

Also... I know what I did wrong. I had 2x^2-2 = 0 on my paper instead of 4.. so I ended up with x^2=1

I need to learn to look over my answers better

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