Okay so I got $x^2 - 5x (p-15)/4 + 27$ for my final answer but I'm sure this is not right. How do I solve and get the right answer for $p$?
I did long division up until I got to the part with $px$, when I got to this part I got $(p-15)x$. I then put $4x$ over $1$ and divided into $(p-15)x$ which in the process canceled out the $x$'s and then I got $(p-15)$.
On
$4x^3-23x^2+px+27=$
$ (4x-3)q(x)$,
where $q(x)$ is a polynomial of degree $2.$
$(4x-3)q(x)=0$ has a root at
$x=3/4.$
Hence
$4(3/4)^3-23(3/4)^2 +p(3/4) +27=0.$
Solve for $p$.
On
You got down to needing $(p-15)x+27$ to be divisible by $4x-3$. As both are linear the quotient is a constant so you have $$(p-15)x+27=k(4x-3)\\(p-15)x+27=4kx-3k$$ To make the constant terms match we must have $k=-9$ so $$p-15=-36\\p=-21$$
On
If you prefer long division we have:
$4x^3 -23x^2 +px +27 \space | \space 4x-3$ which is $x^2-5x+\frac{p-15}{4}$ with the remainder: $\frac{3p-45}{4}+27$. We need the remainder to be zero. So $\frac{3p-45}{4}+27=0$ and $p=-\frac{63}{3}=-21$
On
Alt. hint: let the three roots be $\,a = \dfrac{3}{4}, b, c\,$, then by Vieta's relations:
$\;a+b+c = \dfrac{23}{4} \;\;\implies\;\; b+c = 5$
$\;abc = - \dfrac{27}{4} \;\;\implies\;\; bc = -9$
It follows that $\,\dfrac{p}{4}=ab+ac+bc=a \cdot (b+c)+bc= \dfrac{3}{4} \cdot 5 -9 =\ldots\,$
On
If $4x^3-23x^2+px+27$ is divisible by $4x-3$, the quotient must be of the form $x^2+ax-9$ in order to get the lead and constant coefficients to agree. Expanding
$$(4x-3)(x^2+ax-9)=4x^3+(4a-3)x^2-(36+3a)x+27$$
we see that $4a-3=-23$ implies $a=-5$, from which it follows that
$$p=-(36+3a)=-(36-15)=-21$$
Step $1$: $$4x^3-23x^2+px+27=(4x-3)(?x^2\,+\,?x\,+\,?)$$
Step $2$: Since $4x^3=4x\cdot1x^2$ and $27=-3\cdot(-9)$, $$4x^3-23x^2+px+27=(4x-3)(\color{red}1x^2\,+\,?x\,+\,\color{blue}{-9})$$
Step $3$: $$4x^3-23x^2+px+27=(4x-3)(x^2+\color{green}ux-9)\\\implies -23x^2=4ux^2-3x^2\implies u=-5$$
Step $4$: Expand $(4x-3)(x^2-5x-9)$ to find the coefficient of $x$.