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SAT Math

SAT Math: Advanced Math — Complete Study Guide

Master quadratic equations, polynomials, exponential functions, and rational expressions for the Digital SAT. Detailed explanations, Desmos strategies, and worked examples.

Last updated: February 10, 2026Estimated reading time: 2.5 hours

Introduction

The Advanced Math domain on the Digital SAT focuses on nonlinear relationships and higher-level algebraic reasoning. While Algebra usually feels familiar from earlier courses, Advanced Math asks you to connect several ideas at once: manipulating quadratic and polynomial expressions, analyzing function behavior, solving rational equations with restrictions, and interpreting exponential models in real contexts.

On most SAT forms, you can expect about 13 to 15 questions from Advanced Math. That makes it roughly equal in weight to Algebra. Together, Algebra and Advanced Math usually account for about 60 to 70 percent of all SAT Math questions. This is why students who want major score improvement should prioritize these two domains first.

Advanced Math appears in both Module 1 and Module 2. In Module 1, questions tend to be direct skill checks (for example, factor this quadratic, evaluate an expression, or interpret an exponential model). In Module 2, questions often combine skills and add constraints, such as asking you to choose an equivalent expression, identify a parameter value that changes the number of solutions, or compare two nonlinear models represented in different formats.

Many students consider Advanced Math the hardest SAT domain, and that is often true. However, the good news is that the question patterns are highly predictable. Test writers repeatedly use the same structures:

  1. Convert between equivalent forms of a quadratic or polynomial.
  2. Use specific theorems (factor/remainder/discriminant) to avoid long computation.
  3. Interpret what constants mean in context (initial value, growth factor, vertex, asymptote, restriction).
  4. Check whether candidate solutions are valid, especially in rational equations.

If you train these patterns, Advanced Math becomes manageable and even fast.

Mindset for this domain

You do not need to memorize every possible trick. You need a strong process:

  1. Identify the function family (quadratic, polynomial, exponential, rational).
  2. Choose the most efficient representation (factored, standard, vertex, table, graph).
  3. Solve carefully and verify with structure checks (signs, restrictions, reasonableness).

Strategy tip: On Advanced Math, speed usually comes from recognition, not faster arithmetic. When you see a familiar structure, use the matching tool immediately.


1. Quadratic Equations & Expressions

Core concepts

A quadratic expression has degree 2 and is commonly written in standard form:

ax2+bx+cax^2 + bx + c

A quadratic equation sets that expression equal to zero:

ax2+bx+c=0ax^2 + bx + c = 0

where a0a \neq 0.

SAT questions often move among three key forms.

  1. Standard form

    y=ax2+bx+cy = ax^2 + bx + c
  2. Vertex form

    y=a(xh)2+ky = a(x-h)^2 + k

    The vertex is (h,k)(h, k).

  3. Factored form

    y=a(xr1)(xr2)y = a(x-r_1)(x-r_2)

    The roots (x-intercepts) are x=r1x=r_1 and x=r2x=r_2.

Converting between these forms is essential because each form reveals different information:

  • Standard form is easy for coefficient relationships.
  • Vertex form gives max/min immediately.
  • Factored form gives roots immediately.

Factoring patterns you must know

  1. Greatest common factor (GCF)

    6x2+9x=3x(2x+3)6x^2 + 9x = 3x(2x+3)
  2. Trinomial factoring For x2+bx+cx^2+bx+c, find numbers that multiply to cc and add to bb.

  3. Difference of squares

    a2b2=(a+b)(ab)a^2 - b^2 = (a+b)(a-b)
  4. Perfect square trinomials

    a2+2ab+b2=(a+b)2a^2 + 2ab + b^2 = (a+b)^2 a22ab+b2=(ab)2a^2 - 2ab + b^2 = (a-b)^2

Quadratic formula

If factoring is not easy, use:

x=b±b24ac2ax = \frac{-b \pm \sqrt{b^2-4ac}}{2a}

Discriminant and number of real solutions

The discriminant is:

Δ=b24ac\Delta = b^2 - 4ac
  • If Δ>0\Delta > 0: two distinct real solutions.
  • If Δ=0\Delta = 0: one real solution (double root).
  • If Δ<0\Delta < 0: no real solutions.

Completing the square (method)

To convert ax2+bx+cax^2+bx+c into vertex form, especially when a=1a=1:

  1. Group variable terms.
  2. Move constant if needed.
  3. Add and subtract (b2)2\left(\frac{b}{2}\right)^2.
  4. Rewrite as a square binomial.

Example pattern:

x2+6x+5=(x2+6x+9)9+5=(x+3)24x^2+6x+5 = (x^2+6x+9)-9+5 = (x+3)^2-4

Sum and product of roots

For ax2+bx+c=0ax^2+bx+c=0 with roots r1,r2r_1, r_2:

r1+r2=ba,r1r2=car_1+r_2 = -\frac{b}{a}, \qquad r_1r_2 = \frac{c}{a}

This is useful for parameter questions where you are not asked to compute each root directly.

Worked Example 1: Factoring

Worked Example 1: Factor and solve 2x^2 + 7x - 15 = 0

Solve:

2x2+7x15=02x^2 + 7x - 15 = 0

Method A: Trial-and-error factoring

  1. Multiply leading and constant coefficients:

    2(15)=302\cdot(-15) = -30
  2. Find two integers with product 30-30 and sum 77:

    10 and 310 \text{ and } -3
  3. Split middle term:

    2x2+10x3x15=02x^2 + 10x - 3x - 15 = 0
  4. Group terms:

    (2x2+10x)+(3x15)=0(2x^2+10x) + (-3x-15)=0
  5. Factor each group:

    2x(x+5)3(x+5)=02x(x+5) - 3(x+5)=0
  6. Factor common binomial:

    (2x3)(x+5)=0(2x-3)(x+5)=0
  7. Set each factor equal to zero:

    2x3=0x=322x-3=0 \Rightarrow x=\frac{3}{2} x+5=0x=5x+5=0 \Rightarrow x=-5

Solutions:

x=32, 5\boxed{x=\frac{3}{2},\ -5}

Method B: Quadratic formula (alternative)

For a=2a=2, b=7b=7, c=15c=-15:

x=7±724(2)(15)2(2)x=\frac{-7\pm\sqrt{7^2-4(2)(-15)}}{2(2)}=7±49+1204=7±1694=\frac{-7\pm\sqrt{49+120}}{4}=\frac{-7\pm\sqrt{169}}{4}=7±134=\frac{-7\pm13}{4}

So:

x=64=32,x=204=5x=\frac{6}{4}=\frac{3}{2},\quad x=\frac{-20}{4}=-5

Same answers as factoring.

Worked Example 2: Vertex form and graph interpretation

Worked Example 2: h(t) = -16t^2 + 64t + 5

A ball's height is modeled by:

h(t)=16t2+64t+5h(t) = -16t^2 + 64t + 5

Find the maximum height and when it occurs.

  1. Recognize this is a downward-opening parabola (a=16<0a=-16<0), so the vertex gives maximum height.

  2. Compute vertex time:

    t=b2a=642(16)=2t = -\frac{b}{2a} = -\frac{64}{2(-16)} = 2
  3. Compute height at t=2t=2:

    h(2)=16(22)+64(2)+5=64+128+5=69h(2)=-16(2^2)+64(2)+5=-64+128+5=69

Maximum height is 6969 feet at t=2t=2 seconds.

Converting to vertex form (optional confirmation)

h(t)=16(t24t)+5h(t)=-16(t^2-4t)+5

Complete the square inside:

t24t=(t2)24t^2-4t=(t-2)^2-4

So:

h(t)=16[(t2)24]+5h(t)=-16[(t-2)^2-4]+5=16(t2)2+64+5=-16(t-2)^2+64+5=16(t2)2+69=-16(t-2)^2+69

Vertex form confirms vertex (2,69)(2,69).

Final:

Maximum height =69 feet at t=2 seconds\boxed{\text{Maximum height }=69\text{ feet at }t=2\text{ seconds}}

Worked Example 3: Discriminant

Worked Example 3: For what values of k does x^2 + kx + 9 = 0 have exactly one real solution?

For exactly one real solution, discriminant must be zero:

Δ=b24ac=0\Delta = b^2-4ac = 0

Here a=1a=1, b=kb=k, c=9c=9.

  1. Set discriminant to zero:

    k24(1)(9)=0k^2 - 4(1)(9)=0
  2. Simplify:

    k236=0k^2-36=0
  3. Solve:

    k2=36k=±6k^2=36 \Rightarrow k=\pm6

Answer:

k=6 or k=6\boxed{k=6 \text{ or } k=-6}

Desmos Strategy

Use the built-in Desmos calculator to confirm quadratic work quickly.

  1. Find roots visually

    • Enter y=ax2+bx+cy=ax^2+bx+c.
    • Roots are x-intercepts.
    • Tap intercept points to read exact/approximate values.
  2. Find the vertex

    • Graph the quadratic.
    • The highest/lowest point is the vertex.
    • Click point to read coordinates.
  3. Verify factoring

    • Graph both forms, for example:
      • y=2x2+7x15y=2x^2+7x-15
      • y=(2x3)(x+5)y=(2x-3)(x+5)
    • If they overlap perfectly, factorization is correct.

Strategy tip: In multiple-choice "which equation could represent this graph" questions, compare vertex and opening direction first. That eliminates choices fast.


2. Polynomial Expressions & Equations

Core concepts

A polynomial is an expression with nonnegative integer exponents, such as:

4x3x2+7x94x^3 - x^2 + 7x - 9

SAT Advanced Math tests your ability to manipulate and analyze these expressions efficiently.

Operations with polynomials

  1. Addition/subtraction: combine like terms only.
  2. Multiplication: distribute each term carefully.
  3. Division: use long division or synthetic division when dividing by linear factors.

Long division and synthetic division basics

If dividing by (xa)(x-a), synthetic division is often faster. Keep coefficient placeholders for missing powers.

Remainder theorem

If polynomial f(x)f(x) is divided by (xa)(x-a), the remainder is:

f(a)f(a)

Factor theorem

If f(a)=0f(a)=0, then (xa)(x-a) is a factor of f(x)f(x).

Zeros and factors

  • A zero (root) x=ax=a means f(a)=0f(a)=0.
  • Equivalent factor is (xa)(x-a).

End behavior

For large x|x|, leading term dominates.

  • Even degree, positive leading coefficient: both ends up.
  • Even degree, negative leading coefficient: both ends down.
  • Odd degree, positive leading coefficient: left down/right up.
  • Odd degree, negative leading coefficient: left up/right down.

Worked Example 4: Polynomial multiplication

Worked Example 4: Expand (3x^2 - 2x + 4)(x - 3)

Expand:

(3x22x+4)(x3)(3x^2 - 2x + 4)(x - 3)
  1. Distribute by xx:

    x(3x22x+4)=3x32x2+4xx(3x^2 - 2x + 4)=3x^3-2x^2+4x
  2. Distribute by 3-3:

    3(3x22x+4)=9x2+6x12-3(3x^2 - 2x + 4)=-9x^2+6x-12
  3. Add results:

    3x32x2+4x9x2+6x123x^3-2x^2+4x-9x^2+6x-12
  4. Combine like terms:

    3x311x2+10x123x^3-11x^2+10x-12

Final answer:

3x311x2+10x12\boxed{3x^3-11x^2+10x-12}

Worked Example 5: Remainder theorem

Worked Example 5: p(x)=x^3-4x^2+5x-2, find p(3)

Given:

p(x)=x34x2+5x2p(x)=x^3-4x^2+5x-2
  1. Evaluate at x=3x=3:

    p(3)=334(32)+5(3)2p(3)=3^3-4(3^2)+5(3)-2
  2. Compute:

    =2736+152=27-36+15-2 =4=4

So:

p(3)=4\boxed{p(3)=4}

Interpretation using remainder theorem:

  • When dividing p(x)p(x) by (x3)(x-3), the remainder is 4.
  • Since remainder is not 0, (x3)(x-3) is not a factor.

Strategy tip: If a question asks whether (xa)(x-a) is a factor, you usually only need to compute f(a)f(a), not perform full division.


3. Exponential Functions & Equations

Core concepts

An exponential function has the variable in the exponent.

General model:

f(x)=abxf(x)=a\cdot b^x

where:

  • aa is initial value (value at x=0x=0).
  • bb is growth/decay factor per unit xx.

Growth vs decay

  • If b>1b>1: exponential growth.
  • If 0<b<10<b<1: exponential decay.

Rate relationships:

  • For growth with rate rr, b=1+rb=1+r.
  • For decay with rate rr, b=1rb=1-r.

Compound interest

A=P(1+rn)ntA=P\left(1+\frac{r}{n}\right)^{nt}
  • PP: principal
  • rr: annual rate (decimal)
  • nn: compounds per year
  • tt: years
  • AA: amount after tt years

Doubling time and half-life (conceptual)

  • Doubling time: time needed for quantity to become 2a2a.
  • Half-life: time needed for quantity to become a2\frac{a}{2}.

SAT questions may ask which model grows faster over an interval or how a model parameter changes interpretation.

Linear vs exponential comparison

Linear change adds a constant amount each step. Exponential change multiplies by a constant factor each step.

Example:

  • Linear: +50 each year
  • Exponential: (\times 1.05) each year

Exponential eventually outpaces linear over long intervals.

Worked Example 6: Exponential growth model

Worked Example 6: Bacteria population triples every 4 hours

A bacteria population starts at 500 and triples every 4 hours.

  1. Write P(t)P(t) after tt hours.

  2. Find population after 12 hours.

  3. Initial value:

    P0=500P_0=500
  4. Triple every 4 hours means factor 3 per 4-hour block:

    P(t)=5003t/4P(t)=500\cdot 3^{t/4}
  5. Evaluate at t=12t=12:

    P(12)=500312/4=50033=50027=13,500P(12)=500\cdot 3^{12/4}=500\cdot 3^3=500\cdot27=13{,}500

Answer:

P(t)=5003t/4,P(12)=13,500\boxed{P(t)=500\cdot3^{t/4}},\qquad \boxed{P(12)=13{,}500}

Worked Example 7: Compound interest

Worked Example 7: 2000 dollars at 5% APR compounded monthly for 3 years

Given:

  • P=2000P=2000
  • r=0.05r=0.05
  • n=12n=12
  • t=3t=3

Use:

A=P(1+rn)ntA=P\left(1+\frac{r}{n}\right)^{nt}
  1. Substitute:

    A=2000(1+0.0512)123A=2000\left(1+\frac{0.05}{12}\right)^{12\cdot3}
  2. Simplify inside:

    A=2000(1.004166)36A=2000(1.004166\ldots)^{36}
  3. Approximate:

    A2000(1.1616)2323.20A\approx 2000(1.1616)\approx 2323.20

Approximate amount after 3 years:

2,323.20 dollars (about)\boxed{2{,}323.20\text{ dollars (about)}}

Worked Example 8: Interpreting exponential model

Worked Example 8: Car depreciation model V(t)=32000(0.85)^t

Given:

V(t)=32,000(0.85)tV(t)=32{,}000(0.85)^t

Interpret purchase price and annual depreciation rate.

  1. Initial value at t=0t=0: V(0)=32,000(0.85)0=32,000V(0)=32{,}000(0.85)^0=32{,}000

So purchase price is $32,000.

  1. Decay factor is 0.85, meaning each year value keeps 85%.

  2. Depreciation rate:

    10.85=0.15=15%1-0.85=0.15=15\%

Answer:

  • Purchase price: $32,000
  • Annual depreciation: 15%

Strategy tip: In abta\cdot b^t, the starting value is always aa. The percent change comes from how far bb is from 1.


4. Rational Expressions & Equations

Core concepts

A rational expression is a ratio of polynomials, such as:

x29x2+5x+6\frac{x^2-9}{x^2+5x+6}

You must track restrictions: denominator cannot equal zero.

Simplifying rational expressions

  1. Factor numerator and denominator completely.
  2. Cancel common factors only (not terms connected by + or -).
  3. State excluded values from original denominator.

Adding and subtracting rational expressions

  1. Find LCD.
  2. Rewrite each expression with LCD.
  3. Combine numerators.
  4. Simplify and keep restrictions.

Multiplying/dividing rational expressions

  • Multiply numerators and denominators, then simplify.
  • For division, multiply by reciprocal first.

Solving rational equations

Common method:

  1. Identify excluded values.
  2. Multiply both sides by LCD.
  3. Solve resulting equation.
  4. Check candidate solutions in original equation for extraneous roots.

Extraneous solutions appear because multiplying by expressions involving variable can introduce invalid values.

Worked Example 9: Simplifying rational expression

Worked Example 9: Simplify (x^2 - 9)/(x^2 + 5x + 6)

Simplify:

x29x2+5x+6\frac{x^2-9}{x^2+5x+6}
  1. Factor numerator:

    x29=(x3)(x+3)x^2-9=(x-3)(x+3)
  2. Factor denominator:

    x2+5x+6=(x+2)(x+3)x^2+5x+6=(x+2)(x+3)
  3. Write factored form:

    (x3)(x+3)(x+2)(x+3)\frac{(x-3)(x+3)}{(x+2)(x+3)}
  4. Cancel common factor (x+3)(x+3):

    x3x+2\frac{x-3}{x+2}
  5. State restrictions from original denominator:

    x2, 3x \neq -2,\ -3

Final:

x3x+2,x2,3\boxed{\frac{x-3}{x+2}},\quad x\neq -2,-3

Worked Example 10: Solving rational equation

Worked Example 10: Solve 2/(x-1) + 3/(x+2) = 1

Solve:

2x1+3x+2=1\frac{2}{x-1}+\frac{3}{x+2}=1
  1. Restrictions:

    x1, 2x\neq1,\ -2
  2. LCD is (x1)(x+2)(x-1)(x+2).

  3. Multiply every term by LCD:

    2(x+2)+3(x1)=(x1)(x+2)2(x+2)+3(x-1)=(x-1)(x+2)
  4. Expand left side:

    2x+4+3x3=5x+12x+4+3x-3=5x+1
  5. Expand right side:

    (x1)(x+2)=x2+x2(x-1)(x+2)=x^2+x-2
  6. Set equation:

    5x+1=x2+x25x+1=x^2+x-2
  7. Rearrange:

    0=x24x30=x^2-4x-3
  8. Solve quadratic:

    x=4±16+122=4±282=2±7x=\frac{4\pm\sqrt{16+12}}{2}=\frac{4\pm\sqrt{28}}{2}=2\pm\sqrt{7}
  9. Check restrictions: neither 2+72+\sqrt7 nor 272-\sqrt7 is 1 or -2, so both are valid.

Final solutions:

x=2+7 or x=27\boxed{x=2+\sqrt7 \text{ or } x=2-\sqrt7}

Strategy tip: If a rational equation looks intimidating, write restrictions first. This prevents losing points to invalid roots.


5. Nonlinear Function Analysis

Core concepts

SAT Advanced Math frequently asks you to analyze nonlinear functions by comparing equations, tables, and graphs.

Identify function type quickly

  • Quadratic: ax2+bx+cax^2+bx+c
  • Exponential: abxab^x
  • Rational: ratio of polynomials
  • Higher polynomial: degree 3\ge 3

Transformations

Starting with f(x)f(x):

  • f(x)+kf(x)+k: shift up by kk
  • f(xh)f(x-h): shift right by hh
  • f(x)-f(x): reflect across x-axis
  • f(x)f(-x): reflect across y-axis
  • af(x)a\cdot f(x): vertical stretch/compress by factor a|a|

Key features to interpret

  • Zeros (x-intercepts)
  • Max/min points
  • Intervals of increase/decrease
  • Symmetry
  • End behavior

Compare representations

SAT may give one function by equation and another by table/graph and ask which grows faster over an interval or which has larger output at specific input values.

Composition of functions

For f(g(x))f(g(x)):

  1. Compute inner function g(x)g(x) first.
  2. Substitute result into outer function ff.

Example:

If f(x)=x2+1f(x)=x^2+1 and g(x)=2x3g(x)=2x-3, then

f(g(x))=(2x3)2+1f(g(x))=(2x-3)^2+1

Worked Example 11: Transformations

Worked Example 11: From f(x)=x^2 to g(x)=-(x-3)^2+5

Given:

f(x)=x2f(x)=x^2g(x)=(x3)2+5g(x)=-(x-3)^2+5

Describe transformations and find vertex.

  1. (x3)(x-3) shifts graph right by 3.
  2. Leading negative reflects graph across x-axis.
  3. +5+5 shifts graph up by 5.

Vertex of form a(xh)2+ka(x-h)^2+k is (h,k)(h,k), so:

vertex=(3,5)\text{vertex}=(3,5)

Answer:

  • Right 3, reflect over x-axis, up 5.
  • Vertex: (3,5)\boxed{(3,5)}.

Worked Example 12: Comparing functions

Worked Example 12: Compare rates of change over an interval

Function A is given by table:

xxA(x)A(x)
17
319
531

Function B is given by equation:

B(x)=4x+2B(x)=4x+2

Which function has greater rate of change over x=1x=1 to x=5x=5?

  1. Rate for Function A over interval:

    A(5)A(1)51=3174=244=6\frac{A(5)-A(1)}{5-1}=\frac{31-7}{4}=\frac{24}{4}=6
  2. Rate for Function B is slope of equation:

    44
  3. Compare: 6>46>4.

So Function A has greater rate of change on that interval.

Answer:

Function A\boxed{\text{Function A}}

Strategy tip: When comparing rates, use the same interval for both functions. Do not compare one-point values to slopes.


Quick Reference: Key Formulas

Advanced Math Formula Box

Quadratic formula

x=b±b24ac2ax = \frac{-b \pm \sqrt{b^2-4ac}}{2a}

Discriminant

Δ=b24ac\Delta=b^2-4ac
  • Δ>0\Delta>0: 2 real roots
  • Δ=0\Delta=0: 1 real root
  • Δ<0\Delta<0: 0 real roots

Vertex x-coordinate (from standard form)

x=b2ax = -\frac{b}{2a}

Sum/product of roots for ax2+bx+c=0ax^2+bx+c=0

r1+r2=ba,r1r2=car_1+r_2=-\frac{b}{a}, \qquad r_1r_2=\frac{c}{a}

Exponential growth/decay model

N(t)=N0(1+r)t (growth)N(t)=N_0(1+r)^t \text{ (growth)}N(t)=N0(1r)t (decay)N(t)=N_0(1-r)^t \text{ (decay)}

Compound interest

A=P(1+rn)ntA=P\left(1+\frac{r}{n}\right)^{nt}

Difference of squares

a2b2=(a+b)(ab)a^2-b^2=(a+b)(a-b)

Perfect square trinomials

a2+2ab+b2=(a+b)2a^2+2ab+b^2=(a+b)^2a22ab+b2=(ab)2a^2-2ab+b^2=(a-b)^2

Digital SAT Tips for Advanced Math

  1. Desmos is your best friend: graph equations to locate intersections, roots, and vertices quickly.
  2. For "which could be the equation" questions, compare graph features (intercepts, turning points, asymptotes) instead of expanding every option.
  3. Factoring is usually faster than the quadratic formula when numbers are friendly.
  4. Always test for extraneous solutions in rational equations.
  5. On polynomial items, evaluate at simple values like x=0x=0 or x=1x=1 to eliminate impossible answer choices quickly.
  6. In exponential contexts, identify the base/factor before calculating. Misreading growth factor is a common miss.
  7. If two methods are available, choose the one that reveals the asked quantity directly (for example, vertex form when asked for maximum).

Strategy tip: SAT hard questions often hide simple patterns behind long wording. Translate to symbols early and ignore extra narrative details.


Practice Problems

Problem 1 (Quadratic roots)

Solve:

x29x+20=0x^2-9x+20=0

A) x=4,5x=4,5
B) x=4,5x=-4,-5
C) x=2,10x=2,10
D) x=1,20x=1,20

Problem 2 (Discriminant)

For what value of kk does

2x2+kx+8=02x^2+kx+8=0

have exactly one real solution?

A) k=±8k=\pm 8
B) k=±4k=\pm 4
C) k=±2k=\pm 2
D) k=±16k=\pm 16

Problem 3 (Remainder theorem)

If

f(x)=x32x2+x+7,f(x)=x^3-2x^2+x+7,

what is the remainder when divided by (x2)(x-2)?

A) 1
B) 3
C) 5
D) 9

Problem 4 (Exponential model)

A quantity starts at 1200 and grows by 10% per year. Which function models the quantity after tt years?

A) 1200(0.10)t1200(0.10)^t
B) 1200(1.10)t1200(1.10)^t
C) 1200+0.10t1200+0.10t
D) 1200(10)t1200(10)^t

Problem 5 (Compound interest)

A principal of $1,500 earns 4% annual interest compounded quarterly. Which expression gives the amount after 5 years?

A) 1500(1.04)51500(1.04)^5
B) 1500(1+0.044)201500\left(1+\frac{0.04}{4}\right)^{20}
C) 1500(1+40.04)51500\left(1+\frac{4}{0.04}\right)^5
D) 1500(1+0.044)51500(1+0.04\cdot4)^5

Problem 6 (Rational equation)

Solve:

1x3=2x+1\frac{1}{x-3}=\frac{2}{x+1}

A) x=5x=5
B) x=7x=7
C) x=1x=-1
D) x=3x=3

Problem 7 (Transformation)

If f(x)=x2f(x)=x^2, which describes g(x)=2(x+4)21g(x)=2(x+4)^2-1?

A) Left 4, vertical stretch by 2, down 1
B) Right 4, vertical stretch by 2, down 1
C) Left 4, vertical shrink by 2, up 1
D) Right 4, reflection, up 1

Problem 8 (Polynomial factor theorem)

If

p(x)=x36x2+11x6,p(x)=x^3-6x^2+11x-6,

which statement is true?

A) (x4)(x-4) is a factor
B) (x3)(x-3) is not a factor
C) (x1)(x-1), (x2)(x-2), and (x3)(x-3) are factors
D) The polynomial has no real zeros

Answers and Explanations

1) A
Factor:

x29x+20=(x4)(x5)x^2-9x+20=(x-4)(x-5)

Roots are x=4x=4 and x=5x=5.

2) A
Exactly one real solution means discriminant 0:

k24(2)(8)=0k264=0k=±8k^2-4(2)(8)=0 \Rightarrow k^2-64=0 \Rightarrow k=\pm 8

3) D
Remainder when dividing by (x2)(x-2) is f(2)f(2):

f(2)=88+2+7=9f(2)=8-8+2+7=9

4) B
10% growth means multiply by factor 1.101.10 each year.

5) B
Use compound interest formula with P=1500P=1500, r=0.04r=0.04, n=4n=4, t=5t=5:

A=1500(1+0.044)45A=1500\left(1+\frac{0.04}{4}\right)^{4\cdot5}

6) B
Cross-multiply:

1x3=2x+1x+1=2(x3)\frac{1}{x-3}=\frac{2}{x+1} \Rightarrow x+1=2(x-3)x+1=2x6x=7x+1=2x-6 \Rightarrow x=7

Check restrictions x3,1x\neq3,-1; 77 is valid.

7) A
(x+4)(x+4) shifts left 4, coefficient 2 gives vertical stretch by 2, and 1-1 shifts down 1.

8) C
Test simple roots:

p(1)=16+116=0p(1)=1-6+11-6=0p(2)=824+226=0p(2)=8-24+22-6=0p(3)=2754+336=0p(3)=27-54+33-6=0

So (x1)(x-1), (x2)(x-2), and (x3)(x-3) are factors.


Advanced Math is challenging, but it is also highly learnable because the SAT repeats structures. Focus on representation (standard, factored, vertex), theorem shortcuts (discriminant, remainder/factor), and careful validity checks (restrictions/extraneous roots). With consistent practice, this domain can become one of your biggest score gains.