Find all zeros of $f(x)=128x^3-48x^2+1$ given that one linear factor occurs twice.
let $f(x) $ be equaal to 0
$128x^3-48x^2+1=0,$
$16x^2(8x-3)+1=0,$
trying $x=1/4$
$16/16(2-3)+1=0,$
$1(-1)+1=0=>0=0$
therefore, th zero of$f(x) $is 1/4 and factor is $4x-1$ and the remainder is 0
but after long division method i found that the remainder is not coming to zero it is coming to 5.
please help me how to find all the factors. thanks
A repeated root turns out to be a root of both $f(x) = 0$ and $f'(x) = 0$. That is, it must be a root of:
$$384x^2 - 96x = 0$$ $$4x^2 - x = 0$$ $$x(4x - 1) = 0$$ $$x = 0, \frac{1}{4}$$
Since $f\left(\frac{1}{4}\right) = 0$ and $f'\left(\frac{1}{4}\right) = 0$, it follows that $(4x - 1)$ is a double root of $f(x) = 0$. So we let
$$f(x) = (4x - 1)^2(ax + b)$$
for some real constants $a, b$.
To find $b$, we let $x = 0$. Then,
$$f(0) = b$$
But from the definition of $f(x)$, we have $f(0) = 1$. Hence $b = 1$.
Now to find $a$, we let $x = 1$. Then we have
$$f(1) = 9(a + 1)$$
But from the definition of $f(x)$, we have $f(1) = 81$. Hence it follows that $81 = 9(a + 1)$, which gives $a = 8$. Therefore,
$$f(x) = (4x - 1)^2(8x + 1)$$
You can expand this out to verify that it is true. Maybe you messed up your long division somewhere.
The same idea works generally, see the AC method.
– Bill Dubuque Jul 04 '14 at 13:15