0
$\begingroup$

Suppose $5x - 2$ is a factor of $x^3 - 3x^2 + kx + 15$. Find $k$.

I've tried getting the $x$ value of the factor $5x - 2 = 0$ and got $x= \frac25$ and replaced all the other $x$s with $\frac25$ and set the equation to be equal to $0$. But I don't know if I'm doing right, show me the steps and value of $x$.

$\endgroup$
1
  • $\begingroup$ The factor theorem states that a polynomial $f(x)$ has a factor $(x-k)$ if and only if $f(k)=0$. So yes you are doing right. $\endgroup$ – Vidyanshu Mishra Oct 18 '19 at 19:47
0
$\begingroup$

Your description of your work makes it sound like you're on the right track.

Let $f(x)=x^3-3x^2+kx+15$. If $5x-2$ is a factor of $f(x)$, then by the remainder theorem, we have $f(\frac25)=0$. Thus,

\begin{align} f(x)&=x^3-3x^2+kx+15\\ f(\frac25)=0&=(\frac25)^3-3(\frac25)^2+k(\frac25)+15\\ 0\color{blue}{\cdot125}&=\bigg(\frac8{125}-\frac{12}{25}+\frac25k+15\bigg)\color{blue}{\cdot125}\\ 0&=8-60+50k+1875\\ -1823&=50k\\ k&=-\frac{1823}{50} \end{align}

The moral of the story: don't be alarmed if your final answer is not a number you expected! Rational numbers are real numbers, too!

$\endgroup$
0
$\begingroup$

Yes the method is fine and we obtain

$$\frac8{125} - 3\frac{4}{25} + k\frac25 + 15=0 \iff8-60+50k+1875 \iff k=-\frac{1823}{50}$$

and indeed

$$x^3 - 3x^2 -\frac{1823}{50}x + 15=\frac1{50} (5 x - 2) (10 x^2 - 26 x - 375) $$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.