Divide both sides by \(2\).
\[Px=\frac{2{x}^{4}-13{x}^{3}+30{x}^{2}-28x+8}{2}\]
Divide both sides by \(x\).
\[P=\frac{\frac{2{x}^{4}-13{x}^{3}+30{x}^{2}-28x+8}{2}}{x}\]
Simplify \(\frac{\frac{2{x}^{4}-13{x}^{3}+30{x}^{2}-28x+8}{2}}{x}\) to \(\frac{2{x}^{4}-13{x}^{3}+30{x}^{2}-28x+8}{2x}\).
\[P=\frac{2{x}^{4}-13{x}^{3}+30{x}^{2}-28x+8}{2x}\]
Factor \(2{x}^{4}-13{x}^{3}+30{x}^{2}-28x+8\) using Polynomial Division.
Factor the following.
\[2{x}^{4}-13{x}^{3}+30{x}^{2}-28x+8\]
First, find all factors of the constant term 8.
\[1, 2, 4, 8\]
Try each factor above using the Remainder Theorem.
Substitute 1 into x. Since the result is not 0, x-1 is not a factor..\(2\times {1}^{4}-13\times {1}^{3}+30\times {1}^{2}-28\times 1+8 = -1\)
Substitute -1 into x. Since the result is not 0, x+1 is not a factor..\(2{(-1)}^{4}-13{(-1)}^{3}+30{(-1)}^{2}-28\times -1+8 = 81\)
Substitute 2 into x. Since the result is 0, x-2 is a factor..\(2\times {2}^{4}-13\times {2}^{3}+30\times {2}^{2}-28\times 2+8 = 0\)
\[x-2\]
Polynomial Division: Divide \(2{x}^{4}-13{x}^{3}+30{x}^{2}-28x+8\) by \(x-2\).
| \[2x^3\] | \[-9x^2\] | \[12x\] | \[-4\] | |
| \[x-2\] | \[2x^4\] | \[-13x^3\] | \[30x^2\] | \[-28x\] | \[8\] |
| \[2x^4\] | \[-4x^3\] | | | |
| | \[-9x^3\] | \[30x^2\] | \[-28x\] | \[8\] |
| | \[-9x^3\] | \[18x^2\] | | |
| | | \[12x^2\] | \[-28x\] | \[8\] |
| | | \[12x^2\] | \[-24x\] | |
| | | | \[-4x\] | \[8\] |
| | | | \[-4x\] | \[8\] |
| | | | | \[\] |
Rewrite the expression using the above.
\[2{x}^{3}-9{x}^{2}+12x-4\]
\[P=\frac{(2{x}^{3}-9{x}^{2}+12x-4)(x-2)}{2x}\]
Factor \(2{x}^{3}-9{x}^{2}+12x-4\) using Polynomial Division.
Factor the following.
\[2{x}^{3}-9{x}^{2}+12x-4\]
First, find all factors of the constant term 4.
\[1, 2, 4\]
Try each factor above using the Remainder Theorem.
Substitute 1 into x. Since the result is not 0, x-1 is not a factor..\(2\times {1}^{3}-9\times {1}^{2}+12\times 1-4 = 1\)
Substitute -1 into x. Since the result is not 0, x+1 is not a factor..\(2{(-1)}^{3}-9{(-1)}^{2}+12\times -1-4 = -27\)
Substitute 2 into x. Since the result is 0, x-2 is a factor..\(2\times {2}^{3}-9\times {2}^{2}+12\times 2-4 = 0\)
\[x-2\]
Polynomial Division: Divide \(2{x}^{3}-9{x}^{2}+12x-4\) by \(x-2\).
| \[2x^2\] | \[-5x\] | \[2\] | |
| \[x-2\] | \[2x^3\] | \[-9x^2\] | \[12x\] | \[-4\] |
| \[2x^3\] | \[-4x^2\] | | |
| | \[-5x^2\] | \[12x\] | \[-4\] |
| | \[-5x^2\] | \[10x\] | |
| | | \[2x\] | \[-4\] |
| | | \[2x\] | \[-4\] |
| | | | \[\] |
Rewrite the expression using the above.
\[2{x}^{2}-5x+2\]
\[P=\frac{(2{x}^{2}-5x+2)(x-2)(x-2)}{2x}\]
Split the second term in \(2{x}^{2}-5x+2\) into two terms.
Multiply the coefficient of the first term by the constant term.
\[2\times 2=4\]
Ask: Which two numbers add up to \(-5\) and multiply to \(4\)?
Split \(-5x\) as the sum of \(-x\) and \(-4x\).
\[2{x}^{2}-x-4x+2\]
\[P=\frac{(2{x}^{2}-x-4x+2)(x-2)(x-2)}{2x}\]
Factor out common terms in the first two terms, then in the last two terms.
\[P=\frac{(x(2x-1)-2(2x-1))(x-2)(x-2)}{2x}\]
Factor out the common term \(2x-1\).
\[P=\frac{(2x-1)(x-2)(x-2)(x-2)}{2x}\]
Use Product Rule: \({x}^{a}{x}^{b}={x}^{a+b}\).
\[P=\frac{(2x-1){(x-2)}^{3}}{2x}\]
P=((2*x-1)*(x-2)^3)/(2*x)
