Let's solve it without calculating $p$ or $q$

$\begin{aligned} &f(x) = g(x) (x+4) = (x^2+3x+2)(x+4) \\ \implies &f(-1) =0 \\ \implies &-p+7-14+q=0 \\ \implies &\boxed{p-q=-7} \\ \\ \text{Let } &k(x)=px^3+5x^2-2x-q \\ \implies &k(1)=p+5-2-q=(p-q)+3=-7+3=\boxed{-4} \\ \end{aligned}$

Multiplying $\frac{f(x)}{g(x)}$ by $g(x)$ , we have $f(x)=(x+4)(x^2+3x+2)$ . Because the degree of $x+4$ is $1$ , $x^2+3x+2$ is $2$ , and the degree of $f(x)$ is three, since $1+2=3$ , then $p$ is the multiple of the coefficients of the leading terms of $\frac{f(x)}{g(x)}$ and $g(x)$ , which is $1\times1$ , which is equal to $1$ . Since $q$ has degree $0$ , $q$ is the multiple of the terms that have degree $0$ in both $g(x)$ and $\frac{f(x)}{g(x)}$ , which is $4\times2=8$ , so $q$ is $8$ . Because the output of a function $a(x)$ when $x$ is $1$ is just the sum of of all the coefficients of the terms of the polynomial (because all the $x^n$ terms are just reduced to $1$ ), we have $1+5-2-8=-4$ , so the value of $px^3+5x^2-2x-q$ when $x$ is one is $\boxed{-4}$

At x=1

$\textcolor{crimson}{\frac{f(x)}{g(x)}=\frac{p(1)+5(1)+2(1)+14(1)+q}{(1)+3(1)+2(1)} =(1)+4 }$

This gives us ,

$\textcolor{#0C6AC7}{p+q=9}$ -----(✪)

Now we'll take the functions again and play with 'em a lil,

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$\textcolor{#BA33D6}{p*x^{3}+7*x^{2}+14x+q = x^{3}+7x^{2}+14x+8}$

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Comparing the coefficients of $x^{3}$ and constants,we got to know that

$\textcolor{#0C6AC7}{p=1}$ and $\textcolor{#0C6AC7}{q=8}$

We can adjust the LHS according to our requirements by doing the following

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$\textcolor{#456461}{p*x^{3}+5*x^{2}+2*x^{2}-2x+16x-q+2q=x^{3}+7x^{2}+14x+8}$

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Shifting the unwanted part to RHS we get,

～～～～～～～～～～～～～～～～～～～～～

$\textcolor{#BA33D6}{p*x^{3}+5*x^{2}-2x-q=x^{3}+7x^{2}-2x^{2}+14x-16x+8-2q}$

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Substituting x=1 in the modified version of the equations we get,

～～～～～～～～～～～～～～～～～～～～～

$\textcolor{#E81990}{p*x^{3}+5*x^{2}-2x-q=1+5-2+8-2(8)=-4}$

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Hence $\textcolor{#D61F06}{-4}$ is the correct answer!

Perhaps this is the best shortcut:

Credits to Vinayak Srivastava

Why didn’t I post this earlier?

When $x=-1$ , we have $f(-1)=-p+7-14+q=-p+q-7$ and $f(-1)=g(-1)\times \frac{f(-1)}{g(-1)}=(1-3+2)\times (-1+4)=0.$ $\therefore -p+q-7=0$ .
Solve to get $p-q=-7$ .
$\color{#D61F06} \therefore \text{When }x=1,~px^3+5x^2-2x-q=p+5-2-q=(p-q)+(5-2)=-7+3=-4.$

Method 1: Shortcut:

$px^3 + 5x^2 - 2x - q = f(x) - 2x^2 - 16x -2q = \dfrac{f(x)}{g(x)} \cdot g(x) - 2x^2 - 16x$

When x = 1: $g(x) = (1)^2 + 3(1) + 2 = 1 + 3 + 2 = 6$

$\dfrac{f(x)}{g(x)} = 1 + 4 = 5$

$2x^2 = 2(1)^2 = 2(1) = 2$

$16x = 16(1) = 16$

$\text{Also, since q is a constant, it will be the product of the constants in } g(x), \text{ and } \dfrac{f(x)}{g(x)} \text{(Also because there is no term with the power of } x^{-1} \text{or lower}).$

$\therefore q = 2 \cdot 4 = 8, 2q = 16$

$px^3 + 5x^2 - 2x - q = (5 \cdot 6) - 2 - 16 - 16 = 30 - 34 = \boxed{-4}$

Method 2: Proper method:

$f(x) = \dfrac{f(x)}{g(x)} \cdot g(x) = (x^2+3x+2)(x+4)$

$=(x^2)(x)+(x^2)(4)+(3x)(x)+(3x)(4)+(2)(x)+(2)(4) = x^3+4x^2+3x^2+12x+2x+8 =x^3+7x^2+14x+8$

$\therefore p = 1, q = 8$

$px^3 + 5x^2 - 2x - q = x^3 + 5x^2 - 2x - 8 = (1)^3 + 5(1)^2 - 2(1) - 8 = 1 + 5 - 2 - 8 = 6 - 10 = \boxed{-4}$

f(x) = g(x) * (x + 4) = (x^2+3x+2) * (x+4) = x^3 + 7x^2 + 14x + 8. Comparing with the given expression for f(x), we find p = 1 and q = 8. Then , px^3 + 5x^2 -2x - q = (1)(1)^3 + 5(1)^2 - 2(1) - 8 = 6 - 10 = -4

• $f(x) = g(x) * (x+4) = x^3 + 3x^2 + 2x + 4x^2 + 12x + 8$
• $p x^3 + 7 x^2 + 14 x + q = x^3 + 7 x^2 + 14 x + 8$
• $p = 1, q = 8$
• $1 \cdot 1^3 + 5 \cdot 1^2 - 2 \cdot 1 - 8 = -4$
• QED

The third equation implies that $\begin{aligned} f(x) = g(x)(x+4) = (x^2+3x+2)(x+4) &= x^3+7x^2+14x+8\\ f(x) = px^3 +7x^2+14x+q &= x^3+7x^2+14x+8 \quad\mid -7x²-14x\\[10pt] 1\cdot x^3 + 8 &= p\cdot x^2 + q\iff\; p=1\quad , q=8 \end{aligned}$ Plugging $x=p=1$ and $q=8$ in the equation, we get $\begin{aligned} 1 + 5 -2 -8 = 6-10 = -4. \end{aligned}$

PS: Shoutout to Vinayak Srivastava, your approach is killer!

$\begin{aligned} g(x)&= x²+3x+2\\ &= x²+x+2x+2\\ &= x(x+1)+2(x+1)\\ &= (x+1)(x+2) \end{aligned}$

Given
$\begin{aligned} \frac{p(x)}{g(x)}&= x+4\\ p(x)&= g(x)(x+4)\\ p(x)&= (x+1)(x+2)(x+4) \end{aligned}$

Now the coefficient of x³ is the product of coefficient of each x in p(x)'s factorization, therefore p = 1×1×1 = 1
And q = product of constants in p(x)'s factorization = 1×2×4 = 8

Then we get, px³+5x²-2x-q = x³+5x²-2x-8

Now substitute x = 1, The value becomes 1+5-2-8 = -4

First, we need to find the value of $p$ and $q$ . $f(x)=g(x)\times(x+4)=(x^{2}+3x+2)(x+4)=x^{3}+7x^{2}+14x=8 \rightarrow p=1, q=8$ . Now that we have got the values of $p$ and $q$ , $px^{3}+5x^2-2x-q=1\times1^{3}+5\times1^{2}-2\times1-8=1+5-2-8=\boxed{-4}$

Since we know that $\frac{f(x)}{g(x)}$ = $x+4$ and $f(x)$ = $px^{3}+7x^{2}+14x+q$ and $g(x)$ $=x^{2}+3x+2$ , $\frac{f(x)}{g(x)}$ = $\frac{px^{3}+7x^{2}+14x+q}{x^{2}+3x+2}$ = $x+4$ . We then multiply to get $px^{3}+7x^{2}+14x+q$ = $x^{3}+7x^2+14x+8$ with $\boxed{p=1}$ and $\boxed{q=8}$ . Now we plug in the values of p and q to get $1(1^{3})+5(1^{2})-2(1)-8$ = $1+5-2-8$ = $6-2-8$ = $4-8$ = $\boxed{-4}$

The general form of a cubic polynomial is $ax^3+bx^2+cx+d$ , we also know that multiplying a quadratic polynomial (one of degree $2$ ) with a linear polynomial (one of degree $1$ ) provides us with a cubic polynomial, i.e,

$(a_1x^2+b_1x+c_1)(a_2x+b_2)= ax^3+bx^2+cx+d$

$(a_1a_2)x^3+ (a_1b_2+b_1a_2)x^2+ (c_1a_2+b_1b_2)x+(c_1b_2)= ax^3+bx^2+cx+d$

Where $a= a_1a_2$ , $b=a_1b_2+b_1a_2$ , $c=c_1a_2+b_1b_2$ and $d=c_1b_2$ by observation of coefficients.

We're here given a cubic polynomial $f(x)=px^3+7x^2+14x+q$ , which when divided by a quadratic polynomial $g(x)=x^2+3x+2$ , gets divided totally and gives the quotient, a linear polynomial, $x+4$ , now

$\dfrac{f(x)}{g(x)}= (x+4) \Rightarrow f(x)= g(x)(x+4)$

Thus $f(x)$ is the product of $g(x)$ , a quadratic polynomial, and $(x+4)$ , a linear polynomial, it follows by our above discussion that $p= 1×1=1$ and $q= 4×2=8$ , which gives the third polynomial with no name, $px^3+5x^2-2x-q$ as,

$N(x)= 1x^3+5x^2-2x-8$

We're assigned to find the value of $N(x)$ at $x=1$ , i.e, $N(1)$ , which gives us,

$N(1)= 1(1)^3+5(1)^2-2(1)-8= \boxed{-4}$

Hey Percy, not gonna put down the common solution? (I broke the unwanted streak completely)

Let $p(x)$ be $px^3+5x^2 - 2x -q$

$f(x) = px^3 + 7x^2 + 14x +q \hspace{3 in} \ldots(1)$

$g(x) = x^2 + 3x + 2 \hspace{3.53 in} \ldots (2)$

$\dfrac{f(x)}{g(x)} = x+4$

$\Rightarrow f(x) = (x+4)(g(x))$

$= (x+4)(x^2 + 3x + 2) \hspace{3 in}$ $[$ Using $(2)]$

$= x^3 + 7x^2 + 14x+ 8\hspace{3.35 in} \ldots(3)$

We get eq.(1) = eq.(3)

$px^3 + 7x^2 + 14x +q = x^3 + 7x^2 + 14x+ 8$

By comparing them we get, p = 1 and q = 8

Putting $p = 1$ and $q = 8$ in $p(1)$ we get

$p(1) = 1(1)^3 + 5(1)^2 - 2(1) - 8$ $\rightarrow p(1) = 1+5-2-8$ $\rightarrow p(1) = -4$

$∴$ The answer is $\boxed{-4}$

Simplifying what we have to find as much as we can:

Substitute $x=1$

$px^3+5x^2-2x-q = p+5-2-q = p-q + 3$ .....(eqn1)

So the unknown part is $p - q$ . Consider $f(x) = px^3 + 7x^2 + 14x + q$ We have to find the value of $p-q((-p+q)$ will also work ). In $f(x)$ we cannot change the sign of $q$ . Since any value of $x$ will satisfy given conditions, we have to choose it wisely so that the value of $px^3$ in f(x) will be $-p$ . Simply Logic tells us that the necessary value of $x = -1$ . Also from given it is obvious that $f(x) = g(x) * (x+4)$

Substituting $x = -1$ in $f(x) = g(x) * (x+4)$

$-p+7-14+q = (1-3+2)(-1+4)$

$p-q = -7$

Substituting in eqn1:

-7+3 = $\fbox {-4}$

Alternative (lengthy) method

Though this is Lengthy it is easier to spot (if you know what I mean) and also gives you the values of $p$ and $q$ (which is not necessary but anyways)

$\dfrac{f(x)}{g(x)} = \dfrac{px^3 + 7x^2 + 14x + q }{x^2 + 3x + 2} = x + 4$ .....(eqn1)

Now, this is true for every value of $x$ .

So, let's just put $x = 0$ in (eqn1). (In other words lets just forget the terms with $x$ )

$\frac{q}{2}= 4$

$q = 8$

Substitute $q = 8$ and this time $x = 1$ in (eqn1)

$\dfrac{p + 7 + 14 + 8 }{1 + 3 + 2} = 1 + 4$

$\dfrac{p +29 }{6} = 5$

$p = 30-29$

$p = 1$

$\fbox {p = 1} and \fbox{q = 8}$

Simple calculations show: $px^3 + 5x^2 - 2x - q$ when $x=1; p=1; q=8$ is $\fbox{-4}$

$\because \dfrac{f(x)}{g(x)}=x+4\therefore f(x)=(x+4)g(x)=(x+4)(x^2+3x+2)$ $\Large{Method\space 1}$ $if\space x=-1\Rightarrow f(-1)=(-1+4)((-1)^2-3+2)=0$ $\Rightarrow -p+7-14+q=0$ $\Rightarrow p-q=7-14=-7$ $p\times (1)^3+5\times (1)^2-2\times (1)-q=p+5-2-q$ $\red{p-q}+3=\red{-7}+3=-4$ $\Large{Method\space 2}$ $f(x)=(x+4)g(x)\Rightarrow f(-4)=0$ $\Rightarrow -64p+112-56+q=0$ $\Rightarrow 64p-q=56..........[1]$ $\because g(-2)=0 \space\blue{Just\space put\space and\space find,\space you\space will \space get \space 0}$ $\Rightarrow f(-2)=0$ $\Rightarrow -8p+\cancel{28-28}+q=0$ $\Rightarrow 8p=q..........[2]$ $[1]\space and\space [2]\Rightarrow 64p-8p=56$ $\Rightarrow p=1,q=8$ $\Rightarrow p\times (1)^3+5\times (1)^2-2\times (1)-q=1+5-2-8=6-10=-4$