Video

Thank you! I enjoy making them and helping! =)

Thank you! This comment makes me so happy! I am so glad you got a 790! It is important to practice with the previous problems because they have been repeating on other exams :-o

Planning to post soon! And yes working on August predictions content as well! =) Thank you for your comment and following. =)

Of course! Hope it was useful! =)

Of course! Hope you found it helpful! =)

Thank you!

Glad you found it helpful! =)

Glad it helped! Thanks for your feedback! =)

Hello! Thank you for reaching out and sharing your challenges with reading comprehension. It's great that you're seeking to improve, and I'm here to help! ? =) Let's address your concerns one by one: Understanding Text and Ultimate Meaning 1. Active Reading: While reading, try to engage with the text actively. This means highlighting key points, summarizing paragraphs in your own words, and asking questions about the text as you go. 2. Main Idea and Details: Focus on identifying the main idea of each paragraph and how the details support it. This will help you understand the overall structure and purpose of the text. 3. I think it could also be helpful to build a foundation with reading...if you go to Khan Academy and go through Language Arts courses (perhpas start with grade 9?) www.khanacademy.org/ela/ ...I think this could help build a more solid foundation. Building Vocabulary 1. Context Clues: Guess the meaning of new words from context before looking them up. Flashcards: Try this app out for SAT Specific vocabulary: www.tyrannosaurusprep.com/sat.html Also, watch my vocab video too ;) czcams.com/video/Lque_jQIfK4/video.html Pronunciation and Fluency Phonetic Practice: Use online dictionaries with pronunciation guides. Read Aloud: This improves fluency and comfort with pronunciation. Audible and Podcasts: Listening to audiobooks and educational podcasts helps with pronunciation and comprehension. You're not alone, and with consistent practice and the right strategies, you can improve your reading comprehension and boost your SAT score. Keep up the hard work, and feel free to reach out if you need more guidance!! =) =)

You're welcome! Glad that you found them helpful! =)

Where did you take the exam? What country?

@@epicexamprep so you are saying, different countries are given different questions?

@@sagarbanstola425 It depends on different time zones! Otherwise, students could share questions if they have taken the exam on the same day.

This was on mine. Or at least a similar one.

Those were the 2 that appeared. 🙂

Good luck! I hope it is 1500+ =)

0 is an integer...However, it asks for the smallest POSITIVE integer...and 0 is neither negative nor positive. Also, unrelated to this problem but an interesting number property of 0 that 0 is even. Hope this helps! 🤓

@@epicexamprep Thank you so much

You're welcome! Hope it helps!! =)

Thank you so much! =)

😉

SAME!!!!

Yes! That would also work! There are a couple different methods to solve this question. I just showed the one that I think is the fastest/simplest way to solve. But to explain further on desmos: Input the Equation: For example, for Option D (correct answer), type: y = 3x^2 + 39x + 18b. After typing the equation, simply type b on a new line and Desmos will prompt you to add a slider. Click on "Add Slider." Plot the Vertical Line..since we are told factor x+2b=0, we know that x=-2b, so we can type that into desmos: x=−2b Drag the slider for b and observe the changes in the graph. Check for intersections where the graph of the quadratic equation meets the x-axis at x=−2b. And we see that b=5. *Using a slider helps you explore a range of values and visually identify when the polynomial intersects the x-axis at the required point, confirming the factorization condition.*** ///// Alternatively, I could have also plugged in -2b for x in each of the answer choices and see which one equates to 0 because for a polynomial to be divisible by x+2b, substituting x=−2b into the polynomial should result in zero. Option A: 3x^2 + 9x + 18b Substitute x = -2b: 3(-2b)^2 + 9(-2b) + 18b Simplify: 3(4b^2) - 18b + 18b Result: 12b^2 Set to zero: 12b^2 = 0 Solve: b = 0 Since b must be a positive integer, b=0 doesn't work. Option B: 3x^2 + 24x + 18b Substitute x = -2b: 3(-2b)^2 + 24(-2b) + 18b Simplify: 3(4b^2) - 48b + 18b Result: 12b^2 - 30b Set to zero: 12b^2 - 30b = 0 Factor: b(12b - 30) = 0 Solve: b = 0 or 12b - 30 = 0 12b = 30 b = 30/12 = 2.5 Since b must be a positive integer and this is a non-integer it doesn't work. Option C: 3x^2 + 30x + 18b Substitute x = -2b: 3(-2b)^2 + 30(-2b) + 18b Simplify: 3(4b^2) - 60b + 18b Result: 12b^2 - 42b Set to zero: 12b^2 - 42b = 0 Factor: b(12b - 42) = 0 Solve: b = 0 or 12b - 42 = 0 12b = 42 b = 42/12 = 3.5 Since b must be a positive integer, this one doesn't work. Option D: 3x^2 + 39x + 18b Substitute x = -2b: 3(-2b)^2 + 39(-2b) + 18b Simplify: 3(4b^2) - 78b + 18b Result: 12b^2 - 60b Set to zero: 12b^2 - 60b = 0 Factor: b(12b - 60) = 0 Solve: b = 0 or 12b - 60 = 0 12b = 60 b = 5 Since b = 5 is a positive integer solution, Option D confirms that x + 2b is a factor. I hope this helps! Thanks for your question. =) =)

@owen your method is much faster I just tried it on desmos.

but both ways work fine so rly just pick what you're comfortable with

Of course! I am here to help. Hope the video helps you =)

Hi! That would not be correct because, First you can expand (x + 27)(x - 7): (x + 27)(x - 7) = x^2 - 7x + 27x - 189 = x^2 + 20x - 189 Then, multiply everything by 3x: 3x(x^2 + 20x - 189) = 3x^3 + 60x^2 - 567x The polynomial 3x^3 + 60x^2 - 567x is cubic, which is different from the original quadratic polynomial 3x^2 + 20x - 63. ... For these ones for the "why" behind it, use the steps I outline in the video. BUT if you are in a hurry on the exam, definitely plug it into desmos and check out the x-intercepts! I hope this helps!! =) =)

I cover these ones in the June Predictions video (sum of solutions and product of solutions)...around the 29:16 mark. czcams.com/video/K8oMsv-dBLY/video.html Hope it helps. =)

Wow! Thank you for such a sweet comment! =) ... We do try to get questions that are most similar to what you will potentially see on the exam. Thank you again! =)

haha! I do it subconsciously😅

This is what I eventually did too. I kept dividing 2380 by different numbers to check the sums of the quotient and the divisor. I finally realized that 2380 and 1 give the largest sum.

hahha! Thank you 😂🥰🥰

Hello! I cover more factoring problems in the June SAT Exam Prediction video: czcams.com/video/K8oMsv-dBLY/video.html Alternatively, another great resource is the collegeboard question bank: satsuitequestionbank.collegeboard.org/ Look in "Advanced Math" (Equivalent Expressions)

Hello! Sorry for my delay!! I did not see this comment! Okay...so if I am understanding you correctly, yes that way could also make sense: 34z^14 + bz^7 + 70 So, you are saying z^7 = x. This transforms the polynomial into: 34x^2 + bx + 70 Then we must factor this polynomial such that b is maximized. So we can express the polynomial as a product of two factors: 34x^2 + bx + 70 = (Ax + B)(Cx + D) Expanding the right side, we get: ACx^2 + (AD + BC)x + BD Now we match the coefficients with the original polynomial: The coefficient of x^2 is AC: AC = 34 The coefficient of x is AD + BC: AD + BC = b The constant term is BD: BD = 70 To maximize b, we need to find suitable values of A, C, B, and D that satisfy these conditions. Consider possible factor pairs for 34 and 70: For 34: (A, C) = (1, 34), (2, 17), (17, 2), (34, 1) For 70: (B, D) = (1, 70), (2, 35), (5, 14), (7, 10), (10, 7), (14, 5), (35, 2), (70, 1) To maximize b, we need the combination that produces the highest value. We take A = 34, C = 1 and B = 1, D = 70: AC = 34 * 1 = 34 BD = 1 * 70 = 70 Calculate b: b = AD + BC b = 34 * 70 + 1 * 1 b = 2380 + 1 b = 2381 I think this is what you were asking, right?? Let me know! Hope this made sense!!! =)

@@epicexamprep yeah exactly, thank you for responsee 🥰

You're welcome! Glad you found it useful! =)

Thank you! I hope it helps if you are taking the Digital SAT =)

C doesn't have to be 70 to be a positive integer...it has to be 70 to be the GREATEST positive integer...I will try to explain more here =) : So we expand: (gz^7 + r)(qz^7 + c) = gqz^14 + gcz^7 + rqz^7 + rc Coefficient of z^14: gq = 34 Coefficient of z^7: gc + rq = b Constant term: rc = 70 Find g and q: Since g and q are positive integers such that gq = 34, the possible pairs (g, q) are: (g, q) = (1, 34), (2, 17), (17, 2), (34, 1) Find r and c: Since rc = 70, the pairs (r, c) can be: (1, 70), (2, 35), (5, 14), (7, 10), (10, 7), (14, 5), (35, 2), (70, 1) We need to find which combination of these pairs will give the MAXIMUM value of b. Calculations for Each Pair (g, q): Here I can test each combination of (g, q) with (r, c): For (g, q) = (1, 34): For (r, c) = (1, 70): b = gc + rq = 1 * 70 + 34 * 1 = 70 + 34 = 104 For (r, c) = (2, 35): b = gc + rq = 1 * 35 + 34 * 2 = 35 + 68 = 103 For (r, c) = (5, 14): b = gc + rq = 1 * 14 + 34 * 5 = 14 + 170 = 184 For (r, c) = (7, 10): b = gc + rq = 1 * 10 + 34 * 7 = 10 + 238 = 248 For (r, c) = (10, 7): b = gc + rq = 1 * 7 + 34 * 10 = 7 + 340 = 347 For (r, c) = (14, 5): b = gc + rq = 1 * 5 + 34 * 14 = 5 + 476 = 481 For (r, c) = (35, 2): b = gc + rq = 1 * 2 + 34 * 35 = 2 + 1190 = 1192 For (r, c) = (70, 1): b = gc + rq = 1 * 1 + 34 * 70 = 1 + 2380 = 2381 So, conclusions....we need to find the GREATEST integer value of B...and for that reason c=70 .... I hope this helps to make a bit more sense? =)

​@@epicexamprepmam I'm v. Happy at ur long answer with humbleness ❤

You're welcome! Glad it was informative =)

Great question! Why Assume z^7 + c Instead of pz^7 + c? The choice to express P(z) as z^7 + c simplifies the problem. If we tried pz^7 + c instead, it would introduce another variable p, making the factorization more complex without necessarily providing more useful information. The key is that we are looking for factors that simplify the polynomial, and z^7 + c is a common simplifying assumption that aligns with the degree of the polynomial we have. By assuming z^7 + c as a factor, we streamline the factorization process and find the maximum value of b efficiently. The key is to simplify while ensuring all terms match the original polynomial's structure. It aligns with the structure and coefficients of the original polynomial, ensuring that we correctly factorize and solve for b.....I hope this makes sense!! =) =)

@@epicexamprep If you make the simplification of (z^7 + c) as a factor, then the overall equation becomes (qz^7 + r)(z^7 + c)(p) = 34z^14 + bz^7 +70. The second unknown (p) is required. This question is trying to make sure you know that the possible values for p and q are factors of 34 and the possible values of r and c are factors of 70. You must then realize the greatest value of b is to maximize one of each to get the largest product (largest factor of 34 is 34 and largest factor of 70 is 70. Using 34 *70 will make the second product 1 * 1). Because of this, p does equal 1 (34 * 1=34), but you have to justify it.

You assume p = 1 because you have to make sure the product is the largest. Factoring it to be 35z^7 and 1z^7 is the only way to get the largest product of 34*70. Intuitively you try to get the 2 largest numbers to multiply each after expanding the brackets (the constant terms of both linear factors). And the only way to do this is with 35 and 1 as the coefficients of the z^7 term.

Hello! Yes, a gap year student is eligible to take the SAT exam. There are no specific restrictions on who can take the SAT based on their educational status or age. As long as you meet the registration requirements and pay the necessary fees, you can take the SAT. This applies to high school students, gap year students, and even individuals who have already graduated from high school...hope this answers your question! =)

@@epicexamprep Thank you so much, mam please give some free resources for the SAT exam.

@@Tariqshah-ql6sh Watch all of our videos! They are free! And be sure to do the BlueBook Exams (all 6) as well as the question bank offered on collegeboard.org

Thank you for your comment and insight into the problem. Your approach to finding the smallest and largest values of b by factoring the polynomial using substitution is indeed a useful method! You can substitute x=z^7 into the polynomial 34x^2+bx+70... This helps us convert the original polynomial into a more manageable quadratic form. When we let x=z^7, the z^14 = (z^7)^2=x^2 .... so rewriting the polynomial in terms of x: 34(z^7)^2 + b(z^7)+70 -> 34x^2+bx+70....After the substitution, I want to factor 34x^2+bx+70 so that one of the factors is in the form (qz^7+r).... And then from there I am looking for two binomials that multiply to give 34x^2+bx+70...The factor pairs of the "c" value constant term are: (1,70),(2,35), (5,14), (7,10).....Then test the pairs....for example (34x+70)(x+1) = 34x^2+104x+70...here b=104 (SMALLEST value of b)....the other possibility is (34x+1)(x+70) = 34x^2+2381x+70 .... Where we can also see the correct answer, 2381 for the GREATEST value of b....There are multiple ways to solve, so use the method that comes most natural to you =)

How did we factor 34x²+bx+70 into two factors (34x+70) and (x+1)? Wouldn't the final expression then be 34x²+ 71x+70?

The smallest possible value of b would actually be 103 made up of 35+68. cuz 34*70 = 2380 and 35*68 also equals 2380. Using 34 and 70 in the front and back won’t ensure the smallest value.

Of course! We are here to help =)

Hello! If it is incomplete, you should directly call collegeboard for them to fix your account. If you have Skype and dial this number (collegeboard) +1 (866) 630-9305 it is free of charge! Hope this helps =)

@@epicexamprep Thank you so much. Mam one more question that I am International student, and the College Board has cut $127 from my account. For International student SAT fee is $103. It cuts $24 extra.

@@Tariqshah-ql6sh It says the following on the collegeboard website : The current SAT registration fee is $60, plus a $43 regional fee. For test dates beginning with August 24, 2024, the SAT registration fee is $68, plus a $43 regional fee. It also says that SOME test centers charge a "test center fee" ($24)...so maybe the extra charge is that? Here is the link: satsuite.collegeboard.org/sat/registration/international-testing/fees Hope it helps!! =)

Hello! We have put the simplest ways in this video. This video, in particular, is full of the most challenging questions that you could potentially encounter on the exam and for many of those there is not a simple and fast shortcut. It is imperative to know the concept behind the question and what is being tested, so that you can recognize it and apply it if you see it on your exam. =) ....We are releasing "quick math hacks" video this week, so be sure to check out that! Thanks for you support. =)

Of course! Hope it helps tomorrow!! =)

Good Luck!! =)

Yes, definitely! That is an upcoming video because vertex form of a parabola is a MUST KNOW concept! =)

@@epicexamprep can that be released before the march sat please if possible thanks