Courses
Courses for Kids
Free study material
Offline Centres
More
Store

# How can synthetic division be used to factor a polynomial?

Last updated date: 11th Sep 2024
Total views: 400.2k
Views today: 10.00k
Verified
400.2k+ views
Hint: First of all, use the Rational Zeros Theorem to make the list of all possible rational roots, then test for possible roots using synthetic division and keep testing the roots using the new reduced coefficient.

Complete step by step solution:
Let’s factor the polynomial,
$f\left( x \right) = 4{x^4} - 8{x^3} - 3{x^2} + 7x - 2$.
First, we have to collect the list of all possible rational roots using the “Rational Zeros Theorem”.
To make the list of rational roots and test for all possible roots using synthetic division. Once you find the root, rewrite the original polynomial with the root just you found factored out using the resulting coefficient from the successful synthetic division.
Keep testing roots using the new reduced coefficient coefficients and continuing to factor the polynomial until it is factored entirely into linear equations.
From the above polynomial, we can clearly see that there is a constant term, so the factors of the constant term, 2, are $\pm 1$ and $\pm 2$.
The leading coefficient of, $f\left( x \right) = 4{x^4} - 8{x^3} - 3{x^2} + 7x - 2$ is 4
So, the factors of leading coefficient, 4,are $\pm 1$, $\pm 2$, and $\pm 4$.
So, now we have to divide all the factors of 2 by all factors of 4.
$\bullet$Factors of 2, $\pm 1$ and $\pm 2$.
$\bullet$Factors of 4, $\pm 1$, $\pm 2$, and $\pm 4$.
$\Rightarrow \left\{ {\left. { \pm \dfrac{1}{1}, \pm \dfrac{2}{1}, \pm \dfrac{2}{4}, \pm \dfrac{1}{4}} \right\}} \right.$
After dividing all the factors of 2 by all factors of 4 we get the following list: -
$\Rightarrow \left\{ { \pm 1, \pm 2, \pm \dfrac{1}{2}} \right., \pm \left. {\dfrac{1}{4}} \right\}$
Now, we have to start the testing of the values until we find a root.
$\Rightarrow $$\begin{gathered} 1\left| \!{\underline {\, \begin{gathered} 4\,\,\,\,\,\, - 8\,\,\,\,\,\,\, - 3\,\,\,\,\,\,\,\,7\,\,\,\,\,\, - 2 \\ \,\,\,\,\,\,\,\,\,\,\,\,4\,\,\,\,\,\,\, - 4\,\,\,\, - 7\,\,\,\,\,\,\,\,\,\,0 \\ \end{gathered} \,}} \right. \\ \,\,4\,\,\,\,\,\,\, - 4\,\,\,\,\,\,\, - 7\,\,\,\,\,\,\,0\,\,\,\,\,\,\, - 2 \\ \end{gathered} This was not a root so we tried another. \begin{gathered} \Rightarrow - 1\left| \!{\underline {\, \begin{gathered} 4\,\,\,\,\,\, - 8\,\,\,\,\,\, - 3\,\,\,\,\,\,\,\,\,7\,\,\,\,\,\, - 2 \\ \,\,\,\,\,\,\,\, - 4\,\,\,\,\,\,\,\,12\,\,\,\,\, - 9\,\,\,\,\,\,\,\,\,2 \\ \end{gathered} \,}} \right. \\ \,\,\,\,\,\,\,\,\,\,\,\,\,4\,\,\,\,\, - 12\,\,\,\,\,\,\,\,9\,\,\,\,\,\, - 2\,\,\,\,\,\,\,\,0 \\ \end{gathered} Now we got the root so, this division tells us that we can factor f(x) as follows: \Rightarrow f\left( x \right) = \left( {x + 1} \right)\left( {4{x^3} - 12{x^2} + 9x - 2} \right) Now we can continue the testing of numbers with a synthetic division to find more roots. However, now we try to divide, 4{x^3} - 12{x^2} + 9x - 2 \begin{gathered} \Rightarrow - 2\left| \!{\underline {\, \begin{gathered} 4\,\,\,\,\,\,\, - 12\,\,\,\,\,\,\,\,9\,\,\,\,\,\,\, - 2 \\ \,\,\,\,\,\,\,\,\,\, - 8\,\,\,\,\,\,\,\,\,40\,\,\,\,\, - 98 \\ \end{gathered} \,}} \right. \\ \,\,\,\,\,\,\,\,\,\,\,\,\,4\,\,\,\,\,\,\,\, - 20\,\,\,\,\,\,49\,\,\,\,\, - 100 \\ \end{gathered} This was not a root so we tried another. \begin{gathered} \Rightarrow 2\left| \!{\underline {\, \begin{gathered} 4\,\,\,\,\,\,\, - 12\,\,\,\,\,\,\,\,9\,\,\,\,\,\,\,\, - 2 \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,8\,\,\,\,\,\, - 8\,\,\,\,\,\,\,\,\,\,\,2 \\ \end{gathered} \,}} \right. \\ \,\,\,\,\,\,\,\,\,\,4\,\,\,\,\,\,\,\, - 4\,\,\,\,\,\,\,\,\,\,1\,\,\,\,\,\,\,\,\,\,\,0 \\ \end{gathered} Now, we have found another root! So that it means we can factor, \Rightarrow$$4{x^3} - 12{x^2} + 9x - 2$ into $\left( {x - 2} \right)\left( {4{x^2} - 4x + 1} \right)$
Hence, we can factor $4{x^2} - 4x + 1$ by hand into $\left( {2x - 1} \right)\left( {2x - 1} \right)$ . So our entire polynomial f(x) factors in the following ways,
$\Rightarrow$$4{x^4} - 8{x^3} - 3{x^2} + 7x - 2 = \left( {x + 1} \right)\left( {4{x^3} - 12{x^2} + 9x - 2} \right)$
$\begin{gathered} = \left( {x + 1} \right)\left( {x - 2} \right)\left( {4{x^2} - 4x + 1} \right) \\ = \left( {x + 1} \right)\left( {x - 2} \right)\left( {2x - 1} \right)\left( {2x - 1} \right) \\ \end{gathered}$

Note:
1) The degree of a polynomial f(x) is the largest power of x in the polynomial.
2) The leading term of a polynomial is the term with the highest power of x and the term that determines the degree of the polynomial.
3) The leading coefficient of a polynomial is the coefficient of the leading term.