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Most Missed SAT Math Question on May 2024 Digital SAT exam
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- Опубликовано: 20 май 2024
- The most missed math question on May 2024 SAT exam!!
34z^14+bz^7+70
In the given expression, b is a positive integer. If qz^7+r is a factor of the expression, where q and r are positive integers, what is the greatest possible value of b?
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It's easier to solve the problem by factoring by grouping: Multiply 34 and 70 to get 2380. When you factor, you must find two numbers whose product is 2380 and sum is b. The two numbers that do this while maximizing b are 1 and 2380; therefore, b is 2380 + 1 = 2381.
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.
Also, if they were to ask for the lowest value of b all you have to do is factor like this: make the z^7 equal to x and the espression becomes 34x^2 + bx +70. Therefore, factor it like this (34x+70) (x+1). This will give you the smallest possible value of b which is 104. For the largest value of b switch the 70 and the 1. So, it's really about factoring and placing the factors of the c value in either the first or second patentheses based on how small or big they want the value of b.
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.
Initially, i wanna thank you for making such a benificial videos, afterward i would say to those people who checks the comments before watching the video, guys honestly this is the only source that i see solves approximately the same questions as the SAT
Dont be lazy just checkout all the videos she has posted, believe in me u will all definietly face the same questions on the SAT.
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! =)
Hello. I have two questions regarding this answer.
The first question is how can you be certain that 34z^14+bz^7+70 = (qz^7+r)(z^7+c) but not (qz^7+r)(az^7+c)?
The second question is they never stated that c is an integer. In fact, c can be a fraction like 1/34 or something.
Although 2381 is the correct answer, I need a rigorous solution to this question. Every solution I've seen is mostly based on guesswork.
I understand your concerns about the solutions you've seen, but I'd like to clarify that the solution to this problem is not based on guesswork.... Here's a detailed explanation to show the rigorous approach taken to solve this problem:
To solve the problem, we need to determine the greatest possible value of b such that 34z^14 + bz^7 + 70 can be factored into the form qz^7 + r.
Given the expression:
34z^14 + bz^7 + 70
Question 1: Why 34z^14 + bz^7 + 70 can be expressed as (qz^7 + r)(z^7 + c) but not (qz^7 + r)(az^7 + c)?
To understand why we can assume 34z^14 + bz^7 + 70 = (qz^7 + r)(z^7 + c) and not (qz^7 + r)(az^7 + c), we need to consider the coefficients and the degrees of the terms involved. When expanding (qz^7 + r)(az^7 + c), we would get:
(qz^7 + r)(az^7 + c) = qaz^14 + qcz^7 + raz^7 + rc
Matching coefficients with 34z^14 + bz^7 + 70, we get:
The coefficient of z^14 is qa = 34.
The coefficient of z^7 is qcz^7 + raz^7, which simplifies to q(c + ra)z^7.
The constant term is rc = 70.
For simplicity, if we assume a = 1, the coefficients match directly without introducing additional variables. Therefore, choosing a = 1 simplifies the matching process, which leads to:
34z^14 + bz^7 + 70 = (34z^7 + r)(z^7 + c)
This simplifies the algebraic matching to:
q = 34
r and c such that rc = 70
b = 34c + r
Question 2: Why does c need to be an integer?
To show that c must be an integer, let's examine the constraint rc = 70 more closely. Here, r and c are factors of 70. If c were a fraction, say m/n, then r * (m/n) = 70 would require r to also be a fraction to maintain the product as an integer.
Given that the polynomial is in standard form with integer coefficients, and that qz^7 + r and z^7 + c must produce integer coefficients when multiplied, c should logically be an integer. Introducing fractions would complicate the factorization and potentially invalidate the integer nature of the polynomial coefficients.
This was not shown in video, but to show a step-by-step approach using integers:
Given the polynomial:
34z^14 + bz^7 + 70
We assume it can be factored as: (This is because we need a factorization that maintains the degree of the polynomial and matches the given coefficients).
(34z^7 + r)(z^7 + c)
Step-by-Step Matching Coefficients
Expanding:
(34z^7 + r)(z^7 + c) = 34z^14 + 34cz^7 + rz^7 + rc
Equating Coefficients:
Coefficient of z^14:
34 = 34
Coefficient of z^7:
b = 34c + r
Constant term:
70 = rc
Identifying Integer Pairs
Since rc = 70, we list the integer pairs (r, c):
(1, 70)
(2, 35)
(5, 14)
(7, 10)
(10, 7)
(14, 5)
(35, 2)
(70, 1)
Calculating b for Each Pair
For each pair (r, c), we calculate:
b = 34c + r
(1, 70):
b = 34 * 70 + 1 = 2380 + 1 = 2381
(2, 35):
b = 34 * 35 + 2 = 1190 + 2 = 1192
(5, 14):
b = 34 * 14 + 5 = 476 + 5 = 481
(7, 10):
b = 34 * 10 + 7 = 340 + 7 = 347
(10, 7):
b = 34 * 7 + 10 = 238 + 10 = 248
(14, 5):
b = 34 * 5 + 14 = 170 + 14 = 184
(35, 2):
b = 34 * 2 + 35 = 68 + 35 = 103
(70, 1):
b = 34 * 1 + 70 = 34 + 70 = 104
The maximum value of b is 2381, which occurs when r = 1 and c = 70.
Thus, the greatest possible value of b is 2381.
This method is quite rigorous and follows a structured process, ensuring that the solution is accurate and not based on arbitrary guesses.
I hope this clarifies why the solution is based on a systematic approach rather than guesswork! =)
Can you explain why C has to be 70 again? If they wanted the maximum value for B wouldn't making C something like 10 or 35 make B greater? (it would add +7 or +2 rather than +1 with C being 70) you mentioned making it 70 to keep it from becoming a fraction but again say C was 10 you could just simplify the fraction 70/10 to 7 and avoid it being a fraction? Thank you
Great question! Let's break it down step by step to understand why c = 70 gives the maximum value for b.
Understanding the Expression for b:
The expression for b in terms of c is: b = 34c + 70/c.
Why c = 70?
To maximize b, we need to consider the behavior of the expression 34c + 70/c:
Direct Contribution: The term 34c increases linearly with c.
Inverse Contribution: The term 70/c decreases as c increases.
The goal is to find a balance where the sum 34c + 70/c is maximized.
Evaluating Different Values of c:
Let’s look at a few possible values for c:
If c = 1:
b = 34 * 1 + 70 / 1 = 34 + 70 = 104
If c = 2:
b = 34 * 2 + 70 / 2 = 68 + 35 = 103
If c = 5:
b = 34 * 5 + 70 / 5 = 170 + 14 = 184
If c = 10:
b = 34 * 10 + 70 / 10 = 340 + 7 = 347
If c = 35:
b = 34 * 35 + 70 / 35 = 1190 + 2 = 1192
If c = 70:
b = 34 * 70 + 70 / 70 = 2380 + 1 = 2381
Simplifying Fractions:
While it’s true that 70/c can be simplified, the key point is to find the value of c that maximizes the sum 34c + 70/c. As shown in the examples, smaller values of c (like 1, 2, 5, 10, etc.) do not maximize b as effectively as c = 70.
Why c = 70 Specifically?
When c = 70:
34c contributes significantly to b (since 34 times a large number is very large).
70/c minimizes its contribution to b (since 70 divided by a large number is small).
Thus, c = 70 strikes the best balance by maximizing 34c while keeping 70/c small, leading to the maximum possible b.
I hope this clarifies why c = 70 is the optimal choice for maximizing b! 😊
Let me know if this helps! Thank you for your comment! =)
One can set q = 1 or q = 34 to get the same maximal value of b.
Why are we able to assume that z^7 + c is a factor? Shouldn't we express it as pz^7 + c? What is making the coefficient of z^7 1?
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.
Does any body knows where can I find similar questions to practice from??
Hello! I cover more factoring problems in the June SAT Exam Prediction video: ruclips.net/video/K8oMsv-dBLY/видео.html
Alternatively, another great resource is the collegeboard question bank: satsuitequestionbank.collegeboard.org/
Look in "Advanced Math" (Equivalent Expressions)
Lol. Glad I didn't get this one on my exam. There was one I haven't seen people discuss that was super hard involving sum of the solutions though.
I cover these ones in the June Predictions video (sum of solutions and product of solutions)...around the 29:16 mark. ruclips.net/video/K8oMsv-dBLY/видео.html
Hope it helps. =)