• 500 New SAT Math Problems – Just 19.99 Today Only
    June 30, 2024

    500 New SAT Math Problems

    500 New SAT Math Problems
    Just 19.99 on Amazon

    Hi everyone! The latest edition of 500 New SAT Math Problems is now available in paperback from Amazon. This edition just has been modified from the previous edition to account for the changes on the Digital SAT.

    The paperback is now on sale on Amazon for only $19.99. Note that once the sale ends (by the end of today), the price of this book will go up to $42.99. 

    The promotion has ended. Thanks to everyone who participated. The book is now available at its regular price here: 500 New SAT Math Problems

    If you have any questions, feel free to contact me at steve@SATPrepGet800.com 

    Thank you all for your continued support!

    A Trick For Free Two Day Shipping

    I would like to finish this post with a little trick you can use to get free 2 day shipping on any of the books you decide to purchase without making any additional purchases. If you have never used Amazon Prime you can sign up for a free month using the following link.

    Sign Up For Amazon Prime For Free

    If you have already had a free trial of Amazon Prime you can simply open up a new Amazon account to get a new free trial. It just takes a few minutes! You will need to use a different email address than the one you usually use.

    This next part is very important! After you finish your transaction, go to your Account, select “Manage my prime membership,” and turn off the recurring billing. This way in a month’s time Amazon will not start charging you for the service.

    After shutting off the recurring billing you will still continue to receive the benefit of free 2 day shipping for one month. This means that as long as you use this new Amazon account for your purchases you can do all of your shopping on Amazon for the next month without having to worry about placing minimum orders to get free shipping.

    Just be aware that certain products from outside sellers do not always qualify for free shipping, so please always check over your bill carefully before you check out.

    Well I hope you decide to take advantage of this very special offer, or at the very least I hope you will benefit from my Amazon “free 2 day shipping trick.” Here is the link one more time:

    Sign Up For Amazon Prime For Free

    If you think your friends might be interested in this special offer, please share it with them on Facebook:

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    Thank you all for your continued support!

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  • SAT Math – Scholarly Unicorn Video – Lesson 4, Question 1
    January 17, 2018

    Unicorn - SAT Math

    Level 1 Geometry – Lines and Angles
    Lesson 4, Question 1

     

    This problem is from The Scholarly Unicorn’s SAT Math Advanced Guide. You can find the book on Amazon here: Unicorn SAT Guide

    Scholarly Unicorn's SAT Math Advanced Guide

    If you would like to share additional solutions or you have any questions on this problem, then please post them in the comments below.

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  • SAT Math – Scholarly Unicorn Video – Lesson 12, Question 1
    January 16, 2018

    Unicorn - SAT Math

    Level 1 Geometry – Circles
    Lesson 12, Question 1

     

    This problem is from The Scholarly Unicorn’s SAT Math Advanced Guide. You can find the book on Amazon here: Unicorn SAT Guide

    Scholarly Unicorn's SAT Math Advanced Guide

    If you would like to share additional solutions or you have any questions on this problem, then please post them in the comments below.

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  • SAT Math – Scholarly Unicorn Video – Lesson 3, Question 1

    Unicorn - SAT Math

    Level 1 Problem Solving – Ratios
    Lesson 3, Question 1

     

    This problem is from The Scholarly Unicorn’s SAT Math Advanced Guide. You can find the book on Amazon here: Unicorn SAT Guide

    Scholarly Unicorn's SAT Math Advanced Guide

    If you would like to share additional solutions or you have any questions on this problem, then please post them in the comments below.

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  • SAT Math – Scholarly Unicorn Video – Lesson 2, Question 1
    January 15, 2018

    Unicorn - SAT Math

    Level 1 Passport to Advanced Math – Factoring
    Lesson 2, Question 1

     

    This problem is from The Scholarly Unicorn’s SAT Math Advanced Guide. You can find the book on Amazon here: Unicorn SAT Guide

    Scholarly Unicorn's SAT Math Advanced Guide

    If you would like to share additional solutions or you have any questions on this problem, then please post them in the comments below.

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  • SAT Math – Scholarly Unicorn Video – Lesson 1, Question 1

    Unicorn - SAT Math

    Level 1 Heart of Algebra – Solving Linear Equations
    Lesson 1, Question 1

     

    This problem is from The Scholarly Unicorn’s SAT Math Advanced Guide. You can find the book on Amazon here: Unicorn SAT Guide

    Scholarly Unicorn's SAT Math Advanced Guide

    If you would like to share additional solutions or you have any questions on this problem, then please post them in the comments below.

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  • Solving Linear Equations – Unicorn Puzzle Activity
    January 10, 2018
  • Scholarly Unicorn SAT Math Advanced Guide – Now Available
    January 8, 2018
  • New SAT Math Book with 1000 Problems – Coming Soon
    January 3, 2018
  • Proofreaders Wanted for New SAT Math Book
    December 11, 2017
  • SAT Level 5 PAM Problem with Solution
    December 8, 2017

    SAT Level 5 Passport to Advanced Math Problem
    with Solution

    Today I would like to provide a solution for the difficult Passport to Advanced Math SAT problem I recently posted. Here is the original post: SAT Level 5 PAM Problem

    Level 5 Passport to Advanced Math

    g(x)=x2 + 4x 1
    h(x)=2x3 + 3x2 + x

    The polynomials g and h are defined above. Which of the following polynomials is divisible by 2x 1 ?

    A) k(x) = g(x) h(x)
    B) k(x) = 12g(x) h(x)
    C) k(x) = g(x) 10h(x)
    D) k(x) = 12g(x) 10h(x)

    First recall that the factor theorem says that p(r) = 0  if and only if x – r is a factor of the polynomial p.

    * Solution using the factor theorem: We note that 2x – 1 = 2(x – 1/2), and use r = 1/2. We have

    g(1/2) = (1/2)2 + 4(1/2) – 1 = 1/4 + 2 – 1 = 1/4 + 1 = 5/4

    h(1/2) = 2(1/2)3 + 3(1/2)2 + 1/2 = 2(1/8) + 3(1/4) + 1/2 = 1/4 + 3/4 + 1/2 = 3/2

    Since 12(5/4) – 10(3/2) = 15 – 15 = 0, we see that if k(x) = 12g(x– 10h(x), then k(1/2) = 0, and it follows that 2x – 1 is a factor of k. So, the answer is choice D.

    Note: To use the factor theorem, we need to divide by a linear polynomial of the form x – r. So, we rewrite 2x – 1 as 2(x – 1/2). We see that 2x – 1 is a factor of the polynomial if and only if x – 1/2  is a factor of the polynomial. So, we use r = 1/2.

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