Help me please :) 1. 2x^4+5x^3+x-1 by x^2-2x+1 2. 2x^3+x^2+4 by x-3 Solve using polynomial long division and show work
Do you know how to do polynomial long division?
No
Would you like to learn how to do it?
If you wanna teach me
Alright, let's start with \((2x^4+5x^3+x-1) / (x^2-2x+1) \).
We start with the first terms. \(2x^4/x^2 = ?\)
Could you do this division?
So \(2x^4/x^2 = 2x^2\).
Okay
Can you multiply polynomials? Now we multiply \(2x^2\times (x^2-2x+1)\).
\[ 2x^2\times(x^2-2x+1) = 2x^4-4x^3+2x^2 \]
We subtract this from the thing being divided. \[ 2x^4+5x^3+x-1 - (2x^4-4x^3+2x^2) = 9x^3-2x^2+x-1 \]
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Basically what I've done so far is the first step.
Ooo okay
Since the divisor starts with \(x^2\), and the remaining part starts with \(9x^3\), we can still keep going. Now we do \(9x^3/x^2 = 9x\).
\(9x(x^2-2x+1) = 9x^3 -18x^2+9x\)
\[ 9x^3-2x^2+x-1 - ( 9x^3 -18x^2+9x) = 16x^2-8x-1 \]
It keeps going like this. Does this make sense?
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We can still keep going since the starting parts are \(x^2\) and \(16x^2\).
\(16x^2/x^2=16\). \(16(x^2-2x+1) = 16x^2-32x+16\). \(16x^2-8x+1 - (16x^2-32x+16) = 24x-15 \).
At this point we have to stop, because the starting parts are \(x^2\) and \(24x\). Since the divisor has a larger exponent, we simply accept it is a remainder. Thus we end with \[ \frac{24x-15}{x^2-2x+1} \]. If we add this to the parts we already divided out earlier, the final answer is: \[ 2x^2+9x+16+\frac{24x-15}{x^2-2x+1} \]
The algorithm here is: Step 1: Take the dividend's largest term and divide by the divisors largest term. Step 2: Multiply Step 1's result with the divisor. Step 3: Subtract Step 2's result from the dividend. Continue until the dividend's largest term is lower than the divisors largest term.
I think it would be great if you tried to do question 2 on your own, but I can help if you still don't get it.
Okay thank you so much <33
https://www.mathsisfun.com/algebra/polynomials-division-long.html This website also explains it a bit.
Thank you bb
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