For all real numbers , the expression can be rewritten in the form where , , , and are integers. If is a zero of the polynomial , what is the value of ?

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SAT Equivalent Expressions Question 207 | Free SAT Prep | Aniko

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Question 207·Hard·Equivalent Expressions

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For all real numbers $x \neq -3$ , the expression

\frac{x^3 + p x^2 + q x + 13}{x + 3}

can be rewritten in the form

x^2 + r x + s + \frac{4}{x + 3},

where $p$ , $q$ , $r$ , and $s$ are integers.

If $x = 1$ is a zero of the polynomial $x^3 + p x^2 + q x + 13$ , what is the value of $p + q$ ?

When a rational expression is rewritten as a polynomial plus a fraction, immediately translate that into a polynomial identity by multiplying both sides by the denominator and matching coefficients; this quickly relates the unknown coefficients. Then, if the problem gives you a root like $x=1$ being a zero, plug that value into the polynomial to get a simple equation in the coefficient combination you care about (here, $p+q$ ). Always ask what the question actually requires: sometimes a single substitution for a given root is enough to find a sum like $p+q$ without solving for every individual coefficient.

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Turn the equivalence into a polynomial identity

If two rational expressions are equal for all $x \neq -3$ , then their numerators must be equal after you multiply both sides by $x+3$ . Write an equation that sets $x^3 + p x^2 + q x + 13$ equal to $(x+3)(x^2 + r x + s) + 4$ .

Compare coefficients

Expand $(x+3)(x^2 + r x + s)$ , add the 4, and then match the coefficients of $x^3$ , $x^2$ , $x$ , and the constant term with $x^3 + p x^2 + q x + 13$ . This should let you express $p$ and $q$ in terms of $r$ and $s$ , and quickly find $s$ .

Express p + q in terms of r

Once you have $p = r+3$ and $q$ in terms of $r$ and $s$ , substitute your value of $s$ to get a simple expression for $p+q$ that involves only $r$ .

Use the zero at x = 1

Use the fact that $x=1$ is a zero of $x^3 + p x^2 + q x + 13$ : substitute $x=1$ into the polynomial, set the result equal to 0, and write an equation that involves $p+q$ . Combine this with your expression for $p+q$ from the previous hint.

Desmos Guide

Use a slider to represent p + q

In Desmos, type A = 0 and create a slider for $A$ . Then type the expression f(A) = 1 + A + 13. This represents substituting $x=1$ into the polynomial and letting $A$ stand for $p+q$ .

Adjust the slider to satisfy the zero condition

Move the slider for $A$ until $f(A)$ equals 0. The value of $A$ at that moment is the value of $p+q$ that makes $x=1$ a zero of the polynomial; that is the answer to the question.

Step-by-step Explanation

Write the polynomial identity

Because the two rational expressions are equal for all real $x \neq -3$ , their numerators must be equal:

x^3 + p x^2 + q x + 13 = (x+3)(x^2 + r x + s) + 4.

This means that if we expand the right-hand side, its coefficients must match those of $x^3 + p x^2 + q x + 13$ .

Expand and match coefficients

Expand the product on the right:

\begin{aligned} (x+3)(x^2 + r x + s) &= x^3 + r x^2 + s x + 3x^2 + 3r x + 3s \\ &= x^3 + (r+3)x^2 + (s+3r)x + 3s. \end{aligned}

Now add the $+4$ :

(x+3)(x^2 + r x + s) + 4 = x^3 + (r+3)x^2 + (s+3r)x + (3s+4).

Match this with $x^3 + p x^2 + q x + 13$ :

$p = r+3$
$q = s+3r$
$13 = 3s+4$ .

From the constant terms:

3s + 4 = 13 \Rightarrow 3s = 9 \Rightarrow s = 3.

$p = r + 3$
$q = 3 + 3r$ .

Therefore

p + q = (r+3) + (3+3r) = 4r + 6.

We now have $p+q$ written in terms of $r$ .

Use that x = 1 is a zero

Because $x=1$ is a zero of $x^3 + p x^2 + q x + 13$ , plugging in $x=1$ must give $0$ :

1^3 + p(1)^2 + q(1) + 13 = 0 \Rightarrow 1 + p + q + 13 = 0.

p + q + 14 = 0 \Rightarrow p + q = -14.

From step 2 we also know $p+q = 4r + 6$ , so you could solve $4r+6=-14$ to find $r$ , but for this question we only need $p+q$ , which is

\boxed{-14}.

Strategy

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Desmos Guide

Step-by-step Explanation

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Strategy

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Desmos Guide

Step-by-step Explanation