Suppose $p(x) = 9x^3 - 30x^{2} + 29x - 8 $. If we wish to solve $p(x) = 0$, then we can observe that $x=1$ is a root of $p$. Then, we can write $(x-1)( \dots ) = 0$. How does one find the expression where the dots are in this particular example? And how does one do this in the general case?
$\endgroup$ 33 Answers
$\begingroup$I agree that long division is probably the simplest method of finding out the factors (for a general case)but I usually refrain from using it(if the polynomial doesn't have higher degrees) by adjusting the terms of the polynomials to get a second degree quadratic equation.This is how it is done-
Note that each parenthesis has a factor $(x-1)$ which is taken common from each parenthesis to get a simple expression $$(9x^3-9x^2)+(-21x^2+21x)+(8x-8)=$$ $$9x^2(x-1)-21x(x-1)+8(x-1)=$$
Finally take $(x-1)$ common from the whole expression to get- $$(x-1)(9x^2-21x+8)$$ Now one can easily factor out the quadratic if you want.
$\endgroup$ 1 $\begingroup$$\text{Divide}\;1-1\;\text{into}\;9-30+29-8$
$\begin{array}{rrrr} 9&-21&+8&\\ 9&-30&+29&-8\\ 9&-9&&\\ &-21&+29&\\ &-21&+21&\\ &&8&-8\\ &&8&-8 \end{array}$
$\text{So,}\;9x^2-21x+8=0\;\text{is your reduced equation.}$
$\endgroup$ $\begingroup$Here's an easy-to-remember trick (It is the same basic idea as Freelancer's answer):
If some $x$ is known to be a solution to the equation $mx^3 + nx^2 +ox + p$, then we know that $(x-1)$ is one of the factors of the polynomial. As you say, the third order polynomial can then be written as $(x-1)(...)$. However, we know that the $(...)$ must be a second-order (quadratic) polynomial, since we factored an $x$ out when we pulled out that $(x-1)$. Thus, the $(...)$ is some quadratic with the form $(ax^2 + bx + c)$, such that
$(x-1)(ax^2+bx+c) = mx^3 + nx^2 +ox + p$.
Now we just need to find the values for $a$, $b$, and $c$, find the solutions for that quadratic, and we'll be done.
Let's use your $p(x)=9x^3−30x^2+29x−8$ as an example.
We know $x = 1$ is one of the roots, so we say that $9x^3−30x^2+29x−8 = (x-1)(ax^2+bx+c)$.
Expanding out the right side gives $9x^3−30x^2+29x−8 = (x-1)(ax^2+bx+c) =ax^3+bx^2 + cx -ax^2 - bx - c = ax^3+(b-a)x^2+(c-b)x-c$
The rest is easy.
Finding c
- Looking at the original $p(x)$, $c$ must be $8$, which gives $ax^3+(b-a)x^2+(8-b)x-8$
Finding b
- $b$ is given with some algebra: Looking at $p(x)$, $8-b$ must equal $29$, and $8-b = 29$ gives $b = 8-29 \to b=-21$. That brings us to $ax^3+(-21-a)x^2+(8-(-21))x-8 = ax^3+(-21-a)x^2+(29)x-8$
Finding a
- Just like with $c$, we can look at $p(x)$ to deduce that $a$ must be $9$, which gives $9x^3+(-21-9)x^2+21x-8 = 9x^3+(-30)x^2+21x-8 = 9x^3-30x^2+21x-8$.
Finding $(...)$
- And so, we've found $a$, $b$, and $c$. Substitute these values into the quadratic form for $(x-1)(ax^2 + bx + c)$ and continue from there.
