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3x^4+5x^3+25x^2+45x-18 zeros:3i,-3i. I need help finding the other two zeros
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if those are the two zeros, then one factor is \((x+3i)(x+3i)=x^2+9\)
Yeah I got that part
that means \[3x^4+5x^3+25x^2+45x-18 =(x^2+9)(\text{something})\] find the something by either thinking or by division
i like the think method, it is easier than dividing
for example, the first term has to be \(3x^2\) otherwise you are not going to get \(3x^4\) when you multiply
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the last term has to be \(-2\) so you will get \(-18\) when you multiply
\[3x^4+5x^3+25x^2+45x-18 =(x^2+9)(3x^2+bx-2)\] and all you need is \(b\)
Is the b 5x?
yes
\[3x^2+5x-2\] is the other factor, which in turn factors as \[(3x-1)(x+2)\]
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OMG thank you
yw
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