Hard Problem with Solution from 320 SAT Math Problems Today I’d like to provide you with solutions to the difficult Passport to Advanced Math problem I posted yesterday. This problem is from the new edition of 320 SAT Math Problems. You can click the image of the book above to view the book on Amazon. Here is the problem one more time with two complete solutions. Level 5 Passport to Advanced Math Which of the following could be an equation for the graph shown in the xy-plane above? (A) y = x(x + 2)(1 – x) (B) y = x2(x + 2)(x – 1) (C) y = x(x + 2)2(1 – x) (D) y = x2(x + 2)3(x – 1)2 Solution by process of elimination: The answer choices show us that the answer must be a polynomial. Since there are 3 turning points, the polynomial must have degree at least 4. We can therefore eliminate choice A (this polynomial has degree 3). Since both “ends” of the graph go in the same direction (they both tend to –∞), the polynomial must have even degree. This eliminates choice D (this polynomial has degree 7). Note that we can also use this same reasoning to eliminate choice A. We can eliminate choice B by observing that if we plug a very large value of x into y = x2(x + 2)(x – 1), we will get a large positive value for y. But according to the graph we should get a negative value for y (since the graph is below the x-axis for large x). So the answer is choice C. Notes: (1) A polynomial has the form anxn + an-1 xn-1 + ⋅⋅⋅ + a1x + a0 where a0, a1, … , an are real numbers. For example, x2 + 2x – 35 is a polynomial. The degree of the polynomial is n. In other words, it is the highest power that appears in the expanded form of the polynomial. (2) If a polynomial is in factored form, then we can get the degree of the polynomial by adding the degrees of the factors. For example, the polynomial in choice A has degree 1 + 1 + 1 = 3. Similarly, the polynomials in choices B, C and D have degrees 4, 4 and 7, respectively. (3) The “end behavior” of the graph of a polynomial can tell you whether the polynomial has even or odd degree. If both ends of the graph head in the same direction (both up to ∞, or both down to –∞), then the polynomial must have even degree. If the ends head in different directions (one up and the other down), then the polynomial must have odd degree. The graph shown in this problem is therefore for an even degree polynomial. (4) The degree of a polynomial is at least one more than the number of “turning points” on its graph. Since the graph in this problem has 3 turning points (at x = –2, –1, and approximately 1/2), the degree of the polynomial must be at least 4. (5) Putting notes (3) and (4) together, we see that the possible degrees for the polynomial whose graph is shown are 4, 6, 8,… * Quick solution: Since the graph does not pass through the x-axis at the zero x = –2, it follows that –2 is a zero with even multiplicity. Equivalently, the factor (x + 2) must appear an even number of times (or equivalently, it must have an even power). This eliminates choices A, B and D. So the answer is choice C. Notes: (1) c is a zero of a polynomial p(x) if p(c) = 0. For example, all of the polynomials in the answer choices have the same zeros. They are –2, 0 and 1. (2) p(c) = 0 if and only if x – c is a factor of the polynomial p(x). (3) The multiplicity of the zero x is the degree of the factor x – c. For example, in choice D, 0 and 1 have multiplicity 2, and –2 has multiplicity 3. (4) If a zero c of a polynomial has odd multiplicity, then the graph of the polynomial passes through the x-axis at x = c. From the graph given, we see that 0 and 1 are zeros of the polynomial with odd multiplicity. (5) If a zero c of a polynomial has even multiplicity, then the graph of the polynomial touches the x-axis at x = c, but does not pass through it. From the graph given, we see that –2 is a zero of the polynomial with even multiplicity. (6) It might be tempting to try to plug in the zeros to try to eliminate answer choices in this problem. Unfortunately, all of the answer choices have polynomials with the same zeros. If you liked this post, please share it with your Facebook friends: And don’t forget to check out my social media pages: Comments comments