The expression can be written in the form , where , , and are constants. What is the value of ?

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SAT Equivalent Expressions Question 92 | Free SAT Prep | Aniko

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Question 92·Medium·Equivalent Expressions

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The expression

(3x^{2}-4x-7)\; -\; (5x^{2}+2x-3)\; +\; 6\bigl(x^{2}-x+1\bigr)

can be written in the form $Ax^{2}+Bx+C$ , where $A$ , $B$ , and $C$ are constants. What is the value of $A+B+C$ ?

Your answer

(Express the answer as an integer)

When a polynomial expression is said to equal $Ax^{2}+Bx+C$ and the question asks for $A+B+C$ , recognize that this sum equals the value of the polynomial at $x=1$ . Rather than fully expanding and combining like terms, plug $x=1$ directly into the original expression, compute each piece carefully (watching signs and distribution), and combine the results; this is typically faster and less error-prone on the SAT.

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Think about what $A+B+C$ represents

If the expression is written as $Ax^{2}+Bx+C$ , what do you get when you plug in $x=1$ ? How is that related to $A$ , $B$ , and $C$ ?

Avoid fully expanding if you can

Instead of finding $A$ , $B$ , and $C$ separately, consider evaluating the entire original expression at a single convenient value of $x$ that lets you get $A+B+C$ in one step.

Be careful with signs and parentheses

When you substitute your chosen $x$ -value, handle the minus sign in front of the second parentheses correctly and distribute the $6$ over all three terms in the third parentheses.

Desmos Guide

Enter the expression as a function

In a new line, type:

f(x) = (3x^2 - 4x - 7) - (5x^2 + 2x - 3) + 6(x^2 - x + 1)

Desmos will display the simplified form of this quadratic; from that you can read off $A$ , $B$ , and $C$ if you want.

Use Desmos to find $A+B+C$ directly

In the next line, type f(1). The numerical output is the value of the expression at $x = 1$ , which equals $A+B+C$ for the quadratic $Ax^{2}+Bx+C$ .

Step-by-step Explanation

Relate $A+B+C$ to plugging in a value

Any quadratic written as $Ax^{2}+Bx+C$ has the value

Ax^{2}+Bx+C

at a given $x$ . If you plug in $x=1$ , you get

A(1)^{2}+B(1)+C = A+B+C.

So $A+B+C$ is the value of the expression when $x=1$ .

Substitute $x=1$ into the original expression

Start with

(3x^{2}-4x-7) - (5x^{2}+2x-3) + 6(x^{2}-x+1).

Plug in $x=1$ into each part:

First parentheses: $3(1)^{2} - 4(1) - 7 = 3 - 4 - 7 = -8$ .
Second parentheses: $5(1)^{2} + 2(1) - 3 = 5 + 2 - 3 = 4$ .
Third parentheses: $6(1^{2} - 1 + 1) = 6(1 - 1 + 1) = 6(1) = 6$ .

Now put them together with the original signs:

Start with $-8$ ,
subtract the second result: $-8 - 4 = -12$ ,
add the third result: $-12 + 6 = -6$ .

Connect this value to $A+B+C$ and state the answer

From Step 1, the value of the whole expression at $x=1$ is exactly $A+B+C$ for the quadratic $Ax^{2}+Bx+C$ .

We just found that value at $x=1$ to be $-6$ , so

A+B+C = -6.

Therefore, the answer is $-6$ . (You could also expand and combine like terms to get $4x^{2}-12x+2$ , then compute $4+(-12)+2$ , which also equals $-6$ .)

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