SAT Math - Advanced Math Domain Test

Calculator is allowed on all questions. Figures are not necessarily drawn to scale.

Question 1

Which expression is equivalent to \(x^3\cdot x^5\)?





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When multiplying powers with the same base, add the exponents: \(x^3\cdot x^5=x^{3+5}=x^8\).

Answer: A


Question 2

Which expression is equivalent to \(\sqrt{72}\)?





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\(72=36\cdot2\), so \(\sqrt{72}=\sqrt{36}\sqrt2=6\sqrt2\).

Answer: B


Question 3

Which expression is equivalent to \((x+4)(x-3)\)?





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Distribute: \((x+4)(x-3)=x^2-3x+4x-12=x^2+x-12\).

Answer: A


Question 4

If \(f(x)=2x^2-3\), what is the value of \(f(4)\)?

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\(f(4)=2(4^2)-3=32-3=29\).

Answer: 29


Question 5

Which expression is equivalent to \(x^2-16\)?





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\(x^2-16=x^2-4^2\) is a difference of squares.

it factors as \((x-4)(x+4)\).

Answer: C


Question 6

What are the solutions to \(x^2-9x+20=0\)?





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Factor: \(x^2-9x+20=(x-4)(x-5)\): thus \(x=4\) or \(x=5\).

Answer: C


Question 7

The function \(g\) is defined by \(g(x)=3(2)^x\). What is \(g(2)\)?





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\(g(2)=3(2)^2=3(4)=12\).

Answer: D


Question 8

For \(x\ne3\), the expression \(\frac{x^2-9}{x-3}\) is equivalent to \(x+k\). What is the value of \(k\)?

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Factor the numerator: \(x^2-9=(x-3)(x+3)\).

For \(x\ne3\), the expression simplifies to \(x+3\), so \(k=3\).

Answer: 3


Question 9

Which equation has the same solutions as \(\sqrt{x+5}=x-1\)?





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Squaring both sides gives \(x+5=(x-1)^2\).

Any resulting solution must then be checked in the original equation.

Answer: B


Question 10

If \(p(x)=x^3-4x+7\), what is the remainder when \(p(x)\) is divided by \(x-2\)?





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By the Remainder Theorem, the remainder is \(p(2)=2^3-4(2)+7=8-8+7=7\).

Answer: B


Question 11

The function \(f(x)=x^2+6x+1\) can be written as \((x+3)^2+k\). What is the value of \(k\)?

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\((x+3)^2=x^2+6x+9\): \(x^2+6x+1=(x+3)^2-8\).

Thus \(k=-8\).

Answer: -8


Question 12

If \(f(x)=2x-5\) and \(g(x)=x^2\), which expression is equal to \(g(f(x))\)?





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\(g(f(x))\) means substitute \(f(x)=2x-5\) into \(g(x)=x^2\), giving \((2x-5)^2\).

Answer: C


Question 13

A quadratic function has zeros at \(-2\) and \(6\) and satisfies \(f(0)=-24\). Which equation defines \(f\)?

xy -2 6 (0, -24)



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The zeros give \(f(x)=a(x+2)(x-6)\).

Since \(f(0)=-24\), \(-12a=-24\), so \(a=2\).

Answer: B


Question 14

A population is modeled by \(P(t)=800(1.05)^t\), where \(t\) is the number of years after an initial measurement. What does \(1.05\) represent?

tP 800



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An exponential factor of \(1.05=1+0.05\) represents a 5% increase per year.

Answer: A


Question 15

Selected values of a quadratic function \(f\) are shown in the table. Which equation could define \(f\)?

\(x\) \(f(x)\)
\(-2\) \(7\)
\(0\) \(3\)
\(2\) \(7\)




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The values are symmetric about \(x=0\).

For \(f(x)=x^2+3\), \(f(0)=3\) and \(f(\pm2)=7\), matching the table.

Answer: A


Question 16

If \(\frac{3}{x}+2=5\), what is the value of \(x\)?





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Subtract 2 to get \(\frac{3}{x}=3\): multiplying by \(x\) gives \(3=3x\), so \(x=1\).

Answer: B


Question 17

For the quadratic equation \(x^2+kx+25=0\), there is exactly one real solution. What are the possible values of \(k\)?





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Exactly one real solution occurs when the discriminant is zero: \(k^2-4(1)(25)=0\).

Thus \(k^2=100\), so \(k=-10\) or \(k=10\).

Answer: C


Question 18

The function \(f(x)=a(x-3)^2-5\) satisfies \(f(1)=11\). What is the value of \(a\)?

xy (3, -5) (1, 11)

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Substitute \(x=1\): \(11=a(1-3)^2-5=4a-5\).

Thus \(16=4a\) and \(a=4\).

Answer: 4


Question 19

For \(x>0\), which expression is equivalent to \(\frac{x^{3/2}\sqrt{x^3}}{x}\)?





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For \(x>0\), \(\sqrt{x^3}=x^{3/2}\).

The numerator is \(x^{3/2}x^{3/2}=x^3\), and dividing by \(x\) gives \(x^2\).

Answer: C


Question 20

A function is defined by \(f(x)=\frac{2x+3}{x-4}\). If \(f(a)=3\), what is the value of \(a\)?

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Set \(\frac{2a+3}{a-4}=3\): then \(2a+3=3a-12\), so \(a=15\).

Answer: 15