# Finding the value of k using the factor

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 $$x$$.

• 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. – Vidyanshu Mishra Oct 18 '19 at 19:47

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!

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

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