Posts tagged with: polynomials

This may be a little advanced for the SAT, but complex numbers sometimes show up –as do cubic polynomials– so hopefully you can address this for me! TIA!

Which of the following could be the full set of complex roots of a cubic polynomial with real coefficients?

A. { 0, 1, i}
B. {1, i, 2i}
C. {2, i}
D. {3, 2 + i, 2 – i}

 

I don’t feel like I’m qualified to teach a lesson on this. I know that a cubic polynomial must have 1 or 3 real roots, but I’m too far removed from the class I learned that in to reconstruct the proof here. (A lesson that’s always worth relearning in SAT prep: not all high school math is SAT math.) However, I can still help with this question because it’s multiple choice and happens to lend itself to one of my favorite approaches: we can backsolve!

If you translate the answer choices to what the factors of the polynomial would be, you can multiply those and see which one ends up with all real coefficients. I gravitated right to choice D, because that 2 + i and 2 – i conjugate looks like it’ll cancel out nicely. Let’s see!

We know that a cubic polynomial with roots 3, 2 + i, 2 – i would have the factored form: y=(x-3)(x-(2+i))(x-(2-i)). Now FOIL the complex factors:

y=(x-3)(x-(2+i))(x-(2-i))
y=(x-3)(x^2-(2+i)x-(2-i)x+(2+i)(2-i))
y=(x-3)(x^2-2x-ix-2x+ix+(4-i^2)) <-- All the i terms will go away here!
y=(x-3)(x^2-4x+(4-(-1)))
y=(x-3)(x^2-4x+(4+1))
y=(x-3)(x^2-4x+5)

See how the i terms cancel out? That won’t happen in any other choice, which means the other choices will result in coefficients that aren’t real. We don’t even need to fully expand this one to know it’s the right move.

PWN p. 151 #8 question

In PWN the SAT Math Guide (4th Ed, first printing p. 151), Polynomials chapter question #8, you explain the answer by graphing on calculator. Could you explain the answer to this question without graphing, in particular, why A, C and D are true and why B is false? Many thanks!

Sure! First, recognize that g(x) is factorable:

    \begin{align*}g(x)&=x^2+3x-10\\g(x)&=(x-2)(x+5)\end{align*}

From there, you can see that g(x) has two real zeros (2 and –5) so you can eliminate C.

Then, recognize that g(x) is a factor of f(x) as choice A says:

    \begin{align*}f(x)&=3x^3+9x^2-30x\\f(x)&=3x(x^2+3x-10)\\f(x)&=3x\left(g(x)\right)\end{align*}

Obviously that confirms that choice A is true, so you can eliminate A. However, that also confirms that choice B is false! Remember how we already factored g(x)!

    \begin{align*}f(x)&=3x\left(g(x)\right)\\f(x)&=3x(x-2)(x+5)\end{align*}

f(x) is divisible by (x + 5), but definitely not by (x – 5). You know from the way the question is constructed that only one choice will be false, so once you’ve found it, you can stop working and start bubbling. 🙂

Since you asked, though, we can also show that D is true from what we’ve already figured out using substitution. We already said f(x)=3x\left(g(x)\right), so we can substitute thusly:

    \begin{align*}h(x)&=f(x)+2g(x)\\h(x)&=3x\left(g(x)\right)+2g(x)\end{align*}

We can factor a g(x) out of each of those terms!

    \begin{align*}h(x)&=3x\left(g(x)\right)+2g(x)\\h(x)&=\left(g(x)\right)(3x+2)\end{align*}

Therefore, choice D is also true.

 

f(x) = x^3 – cx^2 + 4x – 4c

In the function f above, c is a constant. How many x-intercepts does the function have?

Can you show how to solve this through logic/algebra? TIA!!

All about factoring here. The giveaway pattern that jumped out to me was the c‘s. Let’s see what we can factor out of first two terms and then what we can factor out of the second two terms.

    \begin{align*}f(x)&=x^3-cx^2+4x-4c\\f(x)&=x^2(x-c)+4(x-c)\\f(x)&=(x^2+4)(x-c)\end{align*}

Once you’ve got that factored, it’s a little easier to see how many x-intercepts there should be. You’ll have an x-intercept for every time you can make the value of the expression zero for some value of x. But can x^2+4 ever equal zero? Nope! So really, the only time the expression will ever equal zero is when x=c. Therefore, there is only one x-intercept for that function.

 

Hi Mike, Can you work the solution for Test 7, Section 3, #13? Thanks!

Absolutely. First of all, this is a great one to plug in on. Equivalent means they’ll be the same for all values of x, so pick a value of x that’s easy to work with and get crackin’. I’m going to use x = 4, because that’ll make the denominator (x – 3) equal to 1.

The original expression:

    \begin{align*}&\dfrac{x^2-2x-5}{x-3}\\\\=&\dfrac{4^2-2(4)-5}{4-3}\\\\=&\dfrac{16-8-5}{1}\\\\=&3\end{align*}

Now, which answer choice gives you 3 when you plug in 4 for x? You can cross off A and B right away because they’re going to be negative—they start with x – 5! So try C and D:

    \begin{align*}\text{C)}\qquad &x+1-\dfrac{8}{x-3}\\\\=&4+1-\dfrac{8}{4-3}\\\\=&-3\\\\\text{D)}\qquad &x+1-\dfrac{2}{x-3}\\\\=&4+1-\dfrac{2}{4-3}\\\\=&3\end{align*}

Boom! D it is.

If you don’t want to plug in, you’re going to need to do polynomial division. (In other words, you should want to plug in!)

Polynomial long division is a real pain to render in text, so please forgive the handwriting.

Step 1: You’re always looking at the lead terms in long division. The leading term of the divisor, x, goes into the leading term of the dividend, x^2, exactly x times. Therefore, put an x at the top, and then subtract the product of x and x-3 from the dividend:

Step 2: x goes into x exactly 1 time, so put a + 1 on top and then subtract the product of x-3 and 1 from the x-5 we’re working with:

Step 3: Now you’re done, because x can’t go into –2. What’s on top is your quotient, and what’s left on bottom is your remainder.

The SAT will often show quotients and remainders as it does in the answer choices of this question: in the \text{QUOTIENT}+\dfrac{\text{REMAINDER}}{\text{DIVISOR}} form. In other words, x+1-\dfrac{2}{x-3} is the answer we’re looking for.

PSAT #2, Section 3, #16

OK, so you’re told that 3x+4 is a factor of 12x^2+ax-20. You need to find a. To do so, you’ll need to set up a classic FOIL scenario. I’m going to use p and r for the unknowns in the second factor.

    \begin{align*}(3x+4)(px+r)&=12x^2+ax-20\end{align*}

Now remember what happens when you FOIL. The 3x and the px will combine to make the 12x^2, and the 4 will combine with the second r to make -20. Therefore, it must be true that p=4 and r=-5. We can use that information to solve for a.

    \begin{align*}(3x+4)(px+r)&=12x^2+ax-20\\(3x+4)(4x-5)&=12x^2+ax-20\\12x^2-15x+16x-20&=12x^2+ax-20\\x&=ax\\1&=a\end{align*}

PSAT #1, Section 3, #17

If x-2 is a factor of x^2-bx+b, then you know that for some other factor x-a,

    \begin{align*}(x-2)(x-a)&=x^2-bx+b\\x^2-2x-ax+2a&=x^2-bx+b\end{align*}

The equivalent polynomials rule tells you that in this case, 2a=b and -2x-ax=-bx. Simplify that second one a bit and you get 2+a=b. You might just see the answer at this point, but you can also solve the system by substituting for b:

    \begin{align*}2+a&=2a\\2&=a\end{align*}

If a=2, then b=4.

Hi Mike,

Can you give me some help with question 10 in the Polynomials section of PWN (pg 149)? I plug in x=-3, per your suggestion, and I get 41 as the answer on the left-hand side of the equation and 3(-3) – 13 + a on the right. That means 41 = -22 + a, or a = 63. What am I missing? Thanks!

Sure! Here’s the question:

The key to this is recognizing that the original equation you’re getting is in quotient-remainder form: 3x-13 is the quotient with a remainder of a. Once you recognize that, you have a few options. You can actually do the division to find the remainder, you can multiply the whole equation by x+3 and solve for a, or you can use the remainder shortcut (since you’re only asked for the remainder in this question) by plugging x=-3 into only the original polynomial.

The shortcut: When a polynomial p(x) is divided by x-k, where k is a constant, you can find the remainder simply by finding p(k). In this case, that means we can find the remainder when 3x^2-4x+2 is divided by x+3 simply by simplifying 3(-3)^2-4(-3)+2, which as you note comes out to 41.

Note that we actually can’t plug in x=-3 in the original equation: we’d get division by zero on both sides.

I’m not going to do the whole long division here, but because I know that shortcut is confusing, I want to stress that just solving the equation for a is also a decent way to go. To begin, multiply everything by x+3.

    \begin{align*}\dfrac{3x^2-4x+2}{x+3}&=3x-13+\dfrac{a}{x+3}\\3x^2-4x+2&=(3x-13)(x+3)+a\\3x^2-4x+2&=3x^2+9x-13x-39+a\\3x^2-4x+2&=3x^2-4x-39+a\\2&=-39+a\\41&=a\end{align*}

Hi Mike. How would you do practise test 8, section 3 question 11? Can you explain the whole polynomials to an odd/even degree etc. The collegeboard answer was a little confusing

Thank you!

Sure. The first thing you need to notice is that the graph has x-intercepts at –3, 0, and 2, which means you need factors of x + 3, x, and x – 2. That lets you eliminate choices C and D. I bet you got that part without much trouble.

After that, you need to choose between choice A, y=x(x-2)(x+3) and choice B, y=x^2(x-2)(x+3). The only difference between the choices, of course, is that one has just x while the other has x^2. Expanded, choice A will begin with an x^3 term, while choice B will begin with an x^4 term.

Basically, you need to know that the highest power of a polynomial tells you whether its y-values will extend to positive and negative infinity, or just to one or the other. If the highest power is odd, the polynomial will go to infinity in both directions, like so:

  

If the power is even, it’ll only go to infinity in one direction, like so:

  

Because the graph in the provided figure only goes towards positive infinity on the y-axis, you know you need the choice that gives you an even powered polynomial. That’s choice B!

 

Problem #25 in the calculator-allowed section of Practice Test #4, please.

This is a fun question! First, note that the polynomials provided have a little something in common:

    \begin{align*}f(x)&=2x^3+6x^2+4x\\g(x)&=x^2+3x+2\end{align*}

Do you see it? How about now:

    \begin{align*}f(x)&=2x(x^2+3x+2)\\g(x)&=x^2+3x+2\\f(x)&=2x[g(x)]\end{align*}

Takeaway lesson from that: when a question at first looks like it might take a very long time to do (in this case, when it looks at first like you’re going to have to do long division possibly four times) slow down for a minute and think about whether there’s a shortcut built into the problem.

You’re looking for a way to combine f(x) and g(x) so that they’re divisible by 2x+3. Once you see that you can pull a 2x (as in, the same 2x that’s part of 2x+3) out of f(x) and get g(x), you know to use that as your base of operations. Let’s just see what happens if we multiply g(x) by 2x+3:

    \begin{align*}&(2x+3)g(x)\\=&2x[g(x)]+3g(x)\end{align*}

Ready for the last step? All we need to do is substitute because we know f(x)=2x\times g(x):

    \begin{align*}&(2x+3)g(x)\\=&2x[g(x)]+3g(x)\\=&f(x)+3g(x)\end{align*}

That matches choice B, and we know that it’s divisible by 2x+3 because we made it by multiplying by 2x+3!

I’m not sure you are answering questions about Practice Test 6 yet. If so, would you please work # 12 in section 3 of that test?

Sure I am—I just haven’t been tracking those answers yet. I’ll get that set up soon.

This one is a straight up plug in question. Say x = 2:

    \begin{align*}&\dfrac{4x^2+6x}{4x+2}\\=&\dfrac{4(2)^2+6(2)}{4(2)+2}\\=&\dfrac{16+12}{8+2}\\=&\dfrac{28}{10}\\=&2.8\end{align*}

Which answer choice simplifies to 2.8?

Obviously not A, B, or C (C is 2 minus something positive, so it can’t be 2.8).

Confirm D:

    \begin{align*}&2+1-\dfrac{2}{4(2)+2}\\=&3-\dfrac{2}{10}\\=&3-0.2\\=&2.8\end{align*}

You could also do the long division…but why? 🙂

That tells you that the quotient is x + 1 with a remainder of 2, just like choice D says.

Would you please show an easy way to solve this:

a(x) = x^3 +3x^2 + 5x
b(x) = 5x^2 + 17x + 16

Polynomial a(x) and b(x) are defined above. Which of the polynomials below has a factor of 3x + 2 ?

(A) l(x) = a(x)+ b(x)
(B) m(x) = 3a(x) + b(x)
(C) n(x) = a(x) – 3b(x)
(D) p(x) = 2a(x) + 3b(x)

This is a ripoff of Official Test 4 Section 4 #25, but it’s a bad ripoff because the original question could be easily factored for a clever and quick solution, and this one can’t. If you get a question like this, you’ll be allowed to use your calculator, though, so the following grind-it-out solution shouldn’t actually take that long.

Note that if a polynomial has a factor of 3x + 2, then it will also have a zero at x=-\dfrac{2}{3}. So the cleverest thing I can think of to do is calculate a\left(-\dfrac{2}{3}\right) and b\left(-\dfrac{2}{3}\right), and then plug those into each answer choice until one gives you zero. Like so:

    \begin{align*}a\left(-\dfrac{2}{3}\right)&=\left(-\dfrac{2}{3}\right)^3+3\left(-\dfrac{2}{3}\right)^2+5\left(-\dfrac{2}{3}\right)=-\dfrac{62}{27}\\b\left(-\dfrac{2}{3}\right)&=5\left(-\dfrac{2}{3}\right)^2+17\left(-\dfrac{2}{3}\right)+16=\dfrac{62}{9}\end{align*}

Now try each answer choice to see which equals zero when a(x)=\dfrac{62}{27} and b(x)=\dfrac{62}{9}. You might be able to see it without much trouble…

A) -\dfrac{62}{27}+\dfrac{62}{9}=\dfrac{124}{27}

B) 3\left(-\dfrac{62}{27}\right)+\dfrac{62}{9}=0

C) -\dfrac{62}{27}-3\left(\dfrac{62}{9}\right)=-\dfrac{620}{27}

D) 2\left(-\dfrac{62}{27}\right)+3\left(\dfrac{62}{9}\right)=\dfrac{434}{27}

Obviously, the answer is B.

Another way to go is to graph each choice with your calculator and look for a zero at x=-\dfrac{2}{3}. That feels to me like it’ll be a bit time consuming, but if you just put your head down and started working on that the minute you saw this question, I bet you’d be done with it within 90 seconds, which is a fine amount of time to spend on a hard question in the calculator section.

 

Hello! Do you think you could explain number 8 in the Polynomials Practice Questions of the PWN the SAT Math Guide 4th Edition? Thank you!

Gladly! I think the best way to go for this one is to factor.

    \begin{align*}f(x)&=3x^3+9x^2-30x\\f(x)&=3x(x^2+3x-10)\\f(x)&=3x(x+5)(x-2)\end{align*}

    \begin{align*}g(x)&=x^2+3x-10\\g(x)&=(x+5)(x-2)\end{align*}

From that, you can see pretty quickly that two answer choices aren’t right. First, you see that g(x) is most definitely a factor of f(x), so A isn’t the answer. You can also see that g(x) has two real zeros (–5 and 2), so C isn’t the answer.

You can also probably see what the answer is. Once everything is factored, it’s clear that (x-5) is not a factor of f(x). That means B is the answer.

As for D, well, once you know the answer is B you might not want to worry about it. But in case it’s keeping you up at night, here’s the deal. If h(x)=f(x)+2g(x), then we can use the factoring we’ve already done to write the following.

    \begin{align*}h(x)&=3x(x+5)(x-2)+2(x+5)(x-2)\end{align*}

See how you can factor (x+5)(x-2) out of both terms?

    \begin{align*}h(x)&=3x[(x+5)(x-2)]+2[(x+5)(x-2)]\\h(x)&=(3x+2)[(x+5)(x-2)]\end{align*}

That’s how you know that (3x+2) is a factor of h(x), which means you can cross off choice D.

can u do test 1 section 3 #15?

Sure! This one is all about corresponding coefficients of equivalent polynomials. When you expand what you’re given…

    \begin{align*}(ax+2)(bx+7)&=15x^2+cx+14\\abx^2+7ax+2bx+14&=15x^2+cx+14\\abx^2+(7a+2b)x+14&=15x^2+cx+14\end{align*}

…you can conclude that the corresponding coefficients of x^2 and x are equal. In other words, you know that ab=15 and 7a+2b=c. Since the question also tells you that a+b=8, you know that either a=3 and b=5 or a=5 and b=3. (Otherwise, ab won’t equal 15!)

The two possible values for c, then, are:

    \begin{align*}7(3)+2(5)&=31\\7(5)+2(3)&=41\end{align*}

That’s answer choice D.

Test 3 Section 3 Number 13 please!

OK, so this is one of those questions that looks way worse than it actually is. Here’s what you’re given:

    \begin{align*}\dfrac{24x^2+25x-47}{ax-2}&=-8x-3-\dfrac{53}{ax-2}\end{align*}

That’s just another way of saying that then the polynomial 24x^2+25x-47 is divided by the polynomial ax-2, the quotient is -8x-3 and the remainder is -53.

Now, here’s the important part. If you think about the first step in polynomial division, it’s this: you divide the leading coefficient of the dividend by the leading coefficient of the divisor. In this case, that means you divide 24 by a. And that tells you the leading coefficient of the quotient, which is –8. What’s the only number you can divide 24 by to get –8? That’s right, a MUST equal –3.

That’s it—you’re done! I’m going to put the full division below just because I like doing math, but in a timed test situation you should stop right there and bubble in B.

t2 sec 3 #17

To get this one, you need to remember that corresponding coefficients of equivalent polynomials are equal. In other words, if ax^2+bx+c is equivalent to mx^2+nx+p then you know that ambn, and cp.

To solve this question, then, all we need to do is simplify the left hand side.

    \begin{align*}2x(3x+5)+3(3x+5)&=ax^2+bx+c\\6x^2+10x+9x+15&=ax^2+bx+c\\6x^2+19x+15&=ax^2+bx+c\end{align*}

From that, we know that a = 6, b = 19, and c = 15. The question only asks us for b, so the answer is 19.