How to factor $2 - 3p - 3p^2 + 2p^3$ to obtain $ (1-2p) \times (1+p) \times (2-p)$ ?

Question

$$2 - 3p - 3p^2 + 2p^3 = (1-2p) \times (1+p) \times (2-p)$$

I want to factor the left hand side to obtain the right hand side.

is there any technique ?

Explain step by step, please.

Answer 1

Yeah thats too much work you guess a number 1 yields 2-3-3+2 thats not 0 how about -1) hey look thats 0 great. that means p+1 is a factor now we know (p+1) divides $2 - 3p - 3p^2 + 2p^3$ so use long division to split it out ie p+1|$2p^3- 3p^2 - 3p +2 $ = $2p^{2}-5p+2$

ie $(2p^{2}-5p+2)(p+1)=2p^3- 3p^2 - 3p +2 $ then complete the squares on the left or go find a textbook that tells u what the quadratic formula is. you can use the method suggest above to guess what values of p may work but don't factor them to try n look for r=0 thats just silly guess them instead.

Especially ina proof where you already know the roots all you need to do is solve for p on 1 of the 3 and then plug that into the LHS once that equal 0 its a root.

Answer 2

$$(1-2p) * (1+p) * (2-p)$$ $$(1-2p)*[(1*2)+(1)(-p)+(p)(2)+(p)(-p)]$$ $$(1-2p)*[2-p+2p-p^2]$$ $$(1-2p)[2+p-p^2]$$ $$[(1*2)+(1*p)+(1)(-p^2)+(-2p*2)+(-2p*p)+(-2p)(-p^2)]$$ $$[2+p-p^2-4p-2p^2+2p^3]$$ $$[2p^3-3p^2-3p+2]$$

2.from LHS=>RHS

$$[2p^3-3p^2-3p+2]$$ $$[2p^3-2p^2-p^2-4p+p+2]$$ $$[2p^3-2p^2-4p-p^2+p+2]$$ $$[(-2p)(-p^2+p+2)+1(-p^2+p+2)]$$ $$[(-2p+1)(-p^2+p+2)]$$ $$[(1-2p)(-p^2+p+2)]$$ $$[(1-2p)(-p^2+2p-p+2)]$$ $$[(1-2p)(-p(p-2)-1(p-2))]$$ $$[(1-2p)(p-2)(-p-1)]$$ $$[-(1-2p)(p-2)(p+1)]$$ $$[(1-2p)(2-p)(p+1)]$$ $$(1-2p)*(1+p)*(2-p)$$

Answer 3

Use Ruffini's rule: the only possible rational factors of the form $p+k$ are those $k$ of the form $\frac{a}{b}$, where $a$ is a factor of the coefficient of the known term and $b$ a factor of the coefficient of the highest-exponent term.

In this case, you should try to divide the left side for $(p+1)$, $(p-1)$, $(p+2)$, $(p-2)$, $(p+\frac{1}{2})$, $(p-\frac{1}{2})$.

Answer 4

The symmetric pattern of coefficients on the left hand side (including signs) tells you that if $p=a$ is a root, then so is $p=\frac 1 a$, so roots come in pairs.

BUT this is a polynomial with odd degree, so one of its roots must stand alone with $a=\frac 1 a$. It is clear that this must be -1 (1 doesn't work). This gets you one root very quickly.

The fact that the roots come in pairs simplifies finding the others using standard methods (or you just have to factorise a quadratic expression)