Function is defined by Function is defined by The graph of in the -plane has -intercepts at , , , and , where , , , and are distinct constants. What is the value of ?

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SAT Nonlinear Functions Question 183 | Free SAT Prep | Aniko

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Question 183·Hard·Nonlinear Functions

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Function $f$ is defined by

f(x)=x^4-14x^3+65x^2-112x

Function $g$ is defined by

g(x)=f(x)+60

The graph of $g$ in the $xy$ -plane has $x$ -intercepts at $(a,0)$ , $(b,0)$ , $(c,0)$ , and $(d,0)$ , where $a$ , $b$ , $c$ , and $d$ are distinct constants. What is the value of $a+b+c+d$ ?

When a question asks for the sum of the roots (or x-intercepts) of a polynomial, do not waste time solving for each root unless necessary. First, write the polynomial in standard form and identify the leading coefficient and the $x^3$ -coefficient. For a monic quartic $x^4+Bx^3+\dots$ , the sum of the roots is $-B$ , so you can immediately get $a+b+c+d$ by taking the negative of that coefficient. Remember that changing only the constant term (like adding $60$ here) does not affect the $x^3$ -coefficient, so the sum of the roots stays tied to that same coefficient.

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Get g(x) in standard polynomial form

First, actually write out $g(x)$ by adding $60$ to $f(x)$ . What is the full expression for $g(x)$ ?

Think about what x-intercepts represent

Each $x$ -intercept $(a,0)$ means that $g(a)=0$ . How can you represent a polynomial if you know its roots (zeros)?

Use the structure of a monic quartic

If a degree-4 polynomial with leading coefficient $1$ factors as $(x-a)(x-b)(x-c)(x-d)$ , what is the coefficient of $x^3$ when you multiply this out? How does that relate to $a+b+c+d$ and to the $x^3$ -coefficient in $g(x)$ ?

Desmos Guide

Enter the function g(x)

In Desmos, type y = x^4 - 14x^3 + 65x^2 - 112x + 60 to graph $g(x)$ .

Find the x-intercepts

Click on or tap the points where the graph crosses the $x$ -axis; Desmos will display the coordinates of these $x$ -intercepts. Note the four $x$ -values you see.

Add the x-intercepts

In a new Desmos expression line, type the sum of the four $x$ -values you recorded (for example, something like x1 + x2 + x3 + x4, using the actual numbers you found). The resulting value should match one of the answer choices.

Step-by-step Explanation

Write g(x) explicitly

We are given

f(x)=x^4-14x^3+65x^2-112x

and

g(x)=f(x)+60.

g(x)=x^4-14x^3+65x^2-112x+60.

This is a degree-4 (quartic) polynomial with leading coefficient $1$ and $x^3$ -coefficient $-14$ .

Connect x-intercepts to roots of g(x)

The $x$ -intercepts $(a,0)$ , $(b,0)$ , $(c,0)$ , and $(d,0)$ are exactly the real zeros of $g(x)$ , so

g(a)=0, \quad g(b)=0, \quad g(c)=0, \quad g(d)=0.

Because $g$ is a degree-4 polynomial with leading coefficient $1$ , it can be factored (in terms of its roots) as

g(x)=(x-a)(x-b)(x-c)(x-d).

Relate the coefficient of x³ to a+b+c+d

Now expand the general product

(x-a)(x-b)(x-c)(x-d).

When you multiply this out, the $x^4$ term is $x^4$ , and the $x^3$ term is

-(a+b+c+d)x^3.

So in any monic quartic of the form $(x-a)(x-b)(x-c)(x-d)$ , the coefficient of $x^3$ is $-(a+b+c+d)$ .

Match coefficients and solve for the sum

We already wrote

g(x)=x^4-14x^3+65x^2-112x+60,

so the coefficient of $x^3$ in $g(x)$ is $-14$ . From the previous step, this coefficient must equal $-(a+b+c+d)$ , so

-(a+b+c+d)=-14.

Multiply both sides by $-1$ to get

a+b+c+d=14.

Thus, the value of $a+b+c+d$ is $14$ , which corresponds to choice B.

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