# Factoring Polynomial: Get Expert Help Now!

• MHB
• bergausstein
In summary: You're right. The factorization you provided is closed form, I thought you were referring to the solution to the polynomial expression. To answer your question, the value of the polynomial expression at $b=\pm\dfrac{1}{2}$ is not zero, so the values of b do not satisfy the polynomial expression.
bergausstein
Please assist me in this problem.$\frac{2}{3}b^5-\frac{1}{6}b^3+\frac{4}{9}b^2-1$

I tried grouping but still could not find anything factorable form of the expression.

Regards.

bergausstein said:
Please assist me in this problem.$\frac{2}{3}b^5-\frac{1}{6}b^3+\frac{4}{9}b^2-1$

I tried grouping but still could not find anything factorable form of the expression.

Regards.

Hi bergausstein,

After thinking it over and cheating a bit (using Wolfram), I'm not sure if a great factorization exists for this polynomial. You can definitely rewrite it a bit, but simpler is relative here. Is this problem from a textbook? What methods are you covering? I'm curious how simplified the answer is intended to be.

As Jameson says, a great factorization is impossible for this polynomial. We can write $$p(x)=(2/3)x^5-(1/6)x^3+(4/9)x^2-1=\frac{1}{18}(\underbrace{12x^5-3x^3+8x^2-18}_{q(x)}).$$ We have $q(1)=\ldots <0$ and $q(2)=\ldots>0,$ so by Bolzano's theorem there is a root $\xi\in (1,2).$
It is not trivial to prove that $q(x)$ has only one real root, but even in this case and using Ruffini algorthim:

$$\begin{array}{r|rrrrrr} & 12 & 0 & -3 & 8 & 0 & -18 \\ \xi & & 12\xi & 12\xi^2 & 12\xi^3-3\xi & 12\xi^4-3\xi^2+8\xi & 12\xi^5-3\xi^3+8\xi^2\\ \hline & 12 & 12\xi & 12\xi^2-3 & 12\xi^3-3\xi+8 & 12\xi^4-3\xi^2+8\xi & \underbrace{12\xi^5-3\xi^3+8\xi^2-18}_{=0} \end{array}$$ So, the factorization of $p(x)$ in $\mathbb{R}[x]$ would be $$p(x)=\dfrac{1}{18}(x-\xi)\left(12x^4+12\xi x^3+(12\xi^2-3)x^2+(12\xi^3-3\xi+8)x+12\xi^4-3\xi^3+8\xi\right).$$ That is, a complete disaster. :)

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As Fernando Revilla pointed out, we can write this polynomial as $$\displaystyle \frac{1}{18}\left(12x^5- 3x^3+ 8x^2- 18\right)$$ so this becomes a question of factoring $$\displaystyle 12x^5- 3x^3+ 8x^2- 18$$.

By the "rational root theorem", if there is a rational root (so that there is a factor with integer coefficients) then it must be of the form $$\frac{m}{n}$$ (and the factor of the form (nx- m))where n is an integer factor of the leading coefficient, 12, and m is an integer factor of the constant term, 18. The integer factors of 12 are 1, -1, 2, -2, 3, -3, 4, -4, 6, -6, 12, and -12 and the integer coefficients of 18 are 1, -1, 2, -2, 3, -3, 6, -6, 9, -9, 18, and -18.

Form all the different fractions with those numerators and denominators (there not as many as there might seem- a lot of cancelling occurs) and put them into the polynomial to see If they work.

bergausstein said:
Please assist me in this problem.$\frac{2}{3}b^5-\frac{1}{6}b^3+\frac{4}{9}b^2-1$

I tried grouping but still could not find anything factorable form of the expression.

Regards.

Short answer, it is not factorizble with real coefficients. Of course every polynomial with complex roots is factorizable with complex coeffcients. I don't know how to get those complex roots but looks like that's not what you are looking for. And what you seem to be looking for is not possible

$\dfrac{2}{3}b^5 - \dfrac{1}{6}b^3+\dfrac{4}{9}b^2 - 1$

$=\dfrac{12}{18}b^5 - \dfrac{3}{18}b^3 + \dfrac{8}{18}b^2 - \dfrac{18}{18}$

$=\dfrac{1}{18}\left(12b^5 - 3b^3 + 8b^2 - 18\right)$

$=\dfrac{1}{18}\left(3b^3(4b^2-1)+2(4b^2-9)\right)$

$=\dfrac{1}{18}\left(3b^3(4b^2-1)+2\left(4b^2-9 \cdot\ \dfrac{4b^2-1}{4b^2-1}\right)\right)$

$=\dfrac{1}{18}\left(3b^3(4b^2-1)+2\left(\dfrac{4b^2-9}{4b^2-1}\right)(4b^2-1)\right)$
$=\dfrac{1}{18}\left((4b^2 - 1)\left(3b^3 + \dfrac{2(4b^2 - 9)}{4b^2 - 1}\right)\right)$
$=\dfrac{1}{18}\left((2b+1)(2b-1)\left(3b^3 + \dfrac{2(2b + 3)(2b-3)}{(2b+1)(2b-1)}\right)\right)$

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bwpbruce said:
$\dfrac{2}{3}b^5 - \dfrac{1}{6}b^3+\dfrac{4}{9}b^2 - 1$
$=\dfrac{1}{18}\left((2b+1)(2b-1)\left(3b^3 + \dfrac{2(2b + 3)(2b-3)}{(2b+1)(2b-1)}\right)\right)$

According to your solution, $b=\pm\dfrac{1}{2}$ are roots. Have you checked them?

Fernando Revilla said:
According to your solution, $b=\pm\dfrac{1}{2}$ are roots. Have you checked them?

Is there not a distinction between factoring a polynomial EXPRESSION and finding the roots of a polynomial EQUATION? Furthermore, even if we were tasked with finding roots, in accordance with the zero product property, $b=\pm\dfrac{1}{2}$ would be $\textit{possible}$ roots. But it is clear that neither value of b would work as a root. It is clear that we have to guess what the root is. If we assume that b = 1, then for the final binomial we have

$3b^3 + \dfrac{2(2b + 3)(2b-3)}{(2b+1)(2b-1)} = 0$
$3(1)^3 + \dfrac{2(2(1) + 3)(2(1)-3)}{(2(1)+1)(2(1)-1)} = 0$

$3(1)^3 + \dfrac{2(5)(-1)}{(3)(1)} = 0$

$3- \dfrac{10}{3} = 0$

$3- 3.33 = 0$

$-(3.33 - 3) = 0$

$-0.33 \approx 0$

So the value of b is close to 1.

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In these problems, we are supposed to provide a closed expression. For example if $p(x)=x^2-2$, then $1.5^2-2=0.25\approx 0.$ A closed expression is $p(x)=(x-\sqrt{2})(x+\sqrt{2})$, not $p(x)=(x-1.5)(x+1.5).$

Fernando Revilla said:
In these problems, we are supposed to provide a closed expression. For example if $p(x)=x^2-2$, then $1.5^2-2=0.25\approx 0.$ A closed expression is $p(x)=(x-\sqrt{2})(x+\sqrt{2})$, not $p(x)=(x-1.5)(x+1.5).$

Which of my expressions are you implying is something other than closed form?

bwpbruce said:
Which of my expressions are you implying is something other than closed form?

(1) You write $$\dfrac{2}{3}b^5 - \dfrac{1}{6}b^3+\dfrac{4}{9}b^2 - 1=\dfrac{1}{18}\left((2b+1)(2b-1)\left(3b^3 + \dfrac{2(2b + 3)(2b-3)}{(2b+1)(2b-1)}\right)\right)$$ What happens when you substitute $b=\pm\dfrac{1}{2}$?

(2) The title of your thread Factoring polynomial, usually means to express a polynomial as a product of irreducible polynomial and does not include rational fractions. Otherwise, you should specify the terms of the factorization you are referring to.

Fernando Revilla said:
(1) You write $$\dfrac{2}{3}b^5 - \dfrac{1}{6}b^3+\dfrac{4}{9}b^2 - 1=\dfrac{1}{18}\left((2b+1)(2b-1)\left(3b^3 + \dfrac{2(2b + 3)(2b-3)}{(2b+1)(2b-1)}\right)\right)$$ What happens when you substitute $b=\pm\dfrac{1}{2}$?

(2) The title of your thread Factoring polynomial, usually means to express a polynomial as a product of irreducible polynomial and does not include rational fractions. Otherwise, you should specify the terms of the factorization you are referring to.

You're still asking about $b=\pm\dfrac{1}{2}$?.

## 1. What is factoring polynomial?

Factoring polynomial is the process of breaking down a polynomial expression into smaller, simpler factors. The goal of factoring is to find the common factors and determine the roots of the equation.

## 2. Why is factoring polynomial important?

Factoring polynomial is important because it helps us solve equations and simplify expressions. It also plays a crucial role in many mathematical concepts, such as finding the zeros of a polynomial function and graphing polynomial equations.

## 3. What are the different methods of factoring polynomial?

There are several methods of factoring polynomial, including GCF (Greatest Common Factor), grouping, difference of squares, trinomial factoring, and perfect square trinomial factoring. Each method is used for specific types of polynomials and has its own set of rules to follow.

## 4. What is the role of factoring polynomial in real life applications?

Factoring polynomial is used in many real-life applications, including finance, engineering, and science. In finance, factoring is used to simplify and solve complex interest and investment problems. In engineering, it is used to solve problems related to optimization and design. In science, factoring is essential for solving equations and modeling real-life situations.

## 5. How can I get expert help with factoring polynomial?

You can get expert help with factoring polynomial by consulting a math tutor or enrolling in a math tutoring program. There are also many online resources, such as videos, practice problems, and forums, that can provide assistance with factoring polynomial. Additionally, many math software programs have built-in tools for factoring polynomial, which can assist with solving equations and simplifying expressions.

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