Polynomials Part2

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Remainder Theorem

When P\left(x\right) is divided by \,x-a, its remainder is P\left(a\right)

Proof

When P\left(x\right) is divided by \,x-a,

  • We know that
    • The quotient is a polynomial (with degree one less than P\left(x\right))
    • The remainder is constant

  • If we use Q\left(x\right) to represent quotient and \,R to represent remainder, we can write an equation
    • P\left(x\right)= \left(x-a\right) Q\left(x\right) \,+R
    • P\left(x\right)=\left(x-a\right)Q\left(x\right)+R is an identity, thus we can substitute any value of \,x
    • The thing is, we wouldn't know what is Q\left(x\right) (unless we do long division), however
      • We see P\left(x\right)={\color{Red}\left(x-a\right)}Q\left(x\right)+R
      • If we can make {\color{Red}\left(x-a\right)} to be zero, then we don't need to worry about Q\left(x\right).
      • To do that, we let \,x=a
  • P\left(x\right)=\left(x-a\right)Q\left(x\right)+R
    • When \,x=a
      • Keep in mind that every \,x must be substituted, thus
        • P\left(x\right) becomes P\left(a\right)
        • Q\left(x\right) becomes Q\left(a\right) (but it's not needed since)
        • \left(x-a\right) becomes zero
    • \therefore P\left(a\right)= \,R
    • Thus, remainder is P\left(a\right)

  • Note: Take some time to admire the beauty of this proof. It involves NO calculations, just some clever steps. The beauty of it will be more apparent when we see how it can be used.

Divided by ax+b

Similarly, we will get

  • P\left(x\right)= \left(ax+b\right)Q\left(x\right)+R
  • Thus, we just choose a value that will make \,ax+b=0
  • For example, if divided by \,3x+2, remainder will be P\left(-\frac{2}{3}\right)
  • Remember In other words, whether it is \,x-a or \,ax+b, just choose a value that will make the divisor zero

Usage

So what's so great about the Remainder Theorem? All it say's is that the remainder is P\left(a\right) when P\left(x\right) is divided by \,x-a.

Well, note that

  • P\left(a\right) is just a value, which we can count easily once we know the value of \,a
  • We don't need to do long division if all we need is the remainder (and not the quotient). For example, we can find remainder when \,x^{4}-3x+7 divided by \,x-1 with a single calculation step instead of (very) long division.

Important Notes

As stated earlier, this subtopic is where HOW you write is as important, or even more important than WHAT you write. Also, make sure you understand WHY it needs to be written that way. For now, keep these two things in mind

  • Must name polynomial (if not given)
  • Must write conclusion

Examples

  • Find remainder when \,x^{2}-3x+8 is divided by \,x-1
  • Let P\left(x\right) = x^{2}-3x+8
    • P\left(1\right) = 6
    • \therefore \mbox{Remainder}\;=6
  • Simple, right? Not exactly. Let's look at some common mistakes

Common Mistakes

Consider the following sample solution

    • x-1 =0, \therefore x=1
    • 1^{2}-3\left(1\right)+8=6
    • \therefore \mbox{Remainder}\;=6
  • If this question was worth 2 marks, personally I will give zero. (The actual STPM scheme might/might not be more lenient, anyway DON'T try your luck)
    • Why? Consider the main working shown, which was 1^{2}-3\left(1\right)+8=6
    • It is a CORRECT statement, but it is not answering the question, and it is not showing that you are using Remainder Theorem
  • Remainder theorem requires the notation of P\left(x\right). Without it, we cannot say that the remainder is P\left(a\right). So any the polynomial must be given a name if it is not already given. We can, of course, use any alphabets, (not necessary capital).

Now consider the following (even more common)

    • Let P\left(x\right) = x^{2}-3x+8
    • P\left(1\right) = 6
  • Can you spot the mistake?
    • There is no conclusion of \mbox{Remainder}\;=6
    • Is it really that important? YES.
      • Without that conclusion, all that is found is P\left(1\right) = 6. So what? It has no meaning by itself.
      • Take note that the conclusion in this case is a step by itself.
        • It is NOT repeating what we have already written
        • It is NOT just to make things neat
      • It it was
        • P\left(x\right) = x^{2}-3x+8
        • \mbox{Remainder} = P\left(1\right) = 6
        • then it is not necessary, cause it was already stated earlier.

Now try doing the next example

  • Find remainder when \,x^{2}-3x+8 is divided by \,3x+2
  • Let P\left(x\right) = x^{2}-3x+8
    • P\left(-\frac{2}{3}\right) = \frac{94}{9}
    • \therefore \mbox{Remainder}\;=\frac{94}{9}

Exercise 4

Find the remainder when P\left(x\right)=x^{2}-5x+6 is divided by the following divisor

Divisor Remainder
\,x-1 P\left(1\right) \,2
\,x+2 P\left(-2\right) \,20
\,x-3 P\left(3\right) \,0
\,2x-1 P\left(\frac{1}{2}\right) \frac{15}{4}
\,3x+2 P\left(-\frac{2}{3}\right) \frac{88}{9}
\,4x-3 P\left(\frac{3}{4}\right) \frac{45}{16}
\,5x+1 P\left(-\frac{1}{5}\right) \frac{176}{25}

Interpreting Information into Equations

Example :

P\left(x\right) gives remainder \,5 when divided by \,x-2

Thus, P\left(2\right)=5

Exercise 5

Complete the following table

Information Equation
P\left(x\right) leaves remainder \,1 when divided by \,x-3 P\left(3\right)=1
When P\left(x\right) is divided by \,x+a its remainder is \,b P\left(-a\right)=b
P\left(x\right) gives remainder \,-20 and \,5 when divided by \,x+2 and \,2x-3 respectively. P\left(-2\right)=-20,P\left(\frac{3}{2}\right)=5
P\left(x\right) is divisible by \,x-4 P\left(4\right)=0
P\left(x\right) have the same remainder when divided by \,2x-3 and \,3x+1 respectively P\left(\frac{3}{2}\right)=P\left(-\frac{1}{3}\right)
P\left(x\right) and G\left(x\right) have the same remainder when divided by \,x-3 P\left(3\right)=G\left(3\right)

Solving Questions

The typical simple question involves

  • Interpreting the given information into equation(s)
  • Solving the equation(s) to find unknowns
  • Checking that the unknown satisfy the information

Simultaneous Equations

When solving simultaneous equation

  • AVOID working with fractions
  • ALWAYS check answers

Getting rid of fractions

  • \frac{1}{4}a+\frac{3}{2}b=2 \Rightarrow \,a+6b=8


  • \frac{8}{27}a-\frac{2}{3}b=-\frac{22}{9} \Rightarrow \,8a-18b=-66


  • \frac{27}{8}a-3b=-\frac{25}{4} \Rightarrow \,27a-24b=-50

Example

\frac{1}{16}a+\frac{1}{2}b=-6 & \,3a-2b=50

  • First, we get rid of the fractions on the first equations.
    • Since the denominators are \,16, 2, we multiply each term with \,16
    • \,a \,+8b \,=-96
  • Since now we have \begin{align} & a+8b=-96 \\ & 3a-2b=50 \end{align} We either multiply one of the equations with an appropriate number or we do substitution. (Multiplication will be shown here)
  • \begin{array}{ll} \frac{1}{16}a+\frac{1}{2}b=-6 & 3a-2b=50 \\ a+8b=-96 \frac{\qquad}{}(1) \qquad \qquad & 12a-8b=200\frac{\qquad}{}(2)\\ (1)+(2) \;:\; 13a = 104 &\\ \therefore a=8 \\ 8+8b=-96 &\\ \therefore b=-13 \\ \therefore a=8 ,b=-13 \\ \end{array}
  • This last step is actually the most important. CHECK!
    • We check by substituting our answers into the original equation(s). Just use calculator to quickly check that satisfies both equations.

More on Checking Solutions

After solving questions, we check that our answers satisfies the original given condition/equation in the question, and not some equations in the middle part.

Example

\,ax^{3}-7x^{2}+bx+7 leaves remainder \,\frac{1}{3} and \,11 when divided by \,3x+2 and \,x-2 respectively. Find the values of \,a and \,b respectively.

Sample solution:

    • Let P\left(x\right)=ax^{3}-7x^{2}+bx+7
    • {\color{Red}\mbox{Step 1 :}} P\left(-\frac{2}{3}\right)=\frac{1}{3} \qquad \qquad \qquad \qquad P\left(2\right)=11
    • {\color{Red}\mbox{Step 2 :}} -\frac{8}{27}+7\left(\frac{4}{9}\right)-\frac{2}{3}b+7=\frac{1}{3} \qquad 8a-28=2b=7=11
    • {\color{Red}\mbox{Step 3 :}} \Bigg\updownarrow
    • {\color{Red}\mbox{Answer :}} a=\frac{3}{8}, b=\frac{29}{2}

Analysis:

  • The fractional values suggest something wrong, though it has to be said that checking must be done even if it is nice round numbers
  • It is NOT enough to substitute the answers into the equations in {\color{Red}\mbox{Step 2}}, we must check at the question
  • To check:
    • Using answers obtained. P\left(x\right)=\frac{3}{8}x^{3}-7x^{2}+\frac{29}{2}x+7
    • We need to check that this polynomial gives the stated remainders.
    • The most definitive check would be doing long division, but of course we are not going to do that.
    • Thus, we do our checking at {\color{Red}\mbox{Step 1}}
      • Note : But before that, we first must check that {\color{Red}\mbox{Step 1}} is correct. If we had done a mistake at {\color{Red}\mbox{Step 1}}, we will get a false-negative from checking (we think there are no mistakes but there is)
    • Using calculator, we get P\left(-\frac{2}{3}\right)=-\frac{53}{9}, \;P\left(2\right)=11 .
    • Thus, we know that there are mistakes in {\color{Red}\mbox{Step 2}} or {\color{Red}\mbox{Step 3}}
    • Our first reaction will normally be check {\color{Red}\mbox{Step 3}}, and in this case it will show that the values indeed to satisfy the equations in {\color{Red}\mbox{Step 2}}, thus we know that the mistake is in {\color{Red}\mbox{Step 2}} itself (it's just a single mistaken \,+ instead of a \,-)
    • In other words, it is not enough to check that we have solve the simultaneous equation correctly ( {\color{Red}\mbox{Step 3}} ), but that the equation itself is ( {\color{Red}\mbox{Step 2}} ) is correct. Checking at the question will almost certainly identify whether there are mistakes at any step.

General Guidelines:

  • Check at the question. If not possible / feasible, check as high as possible.
  • Check that the preceding steps if we are not checking directly at question.
  • If answers are not correct, and we need to narrow down the area of mistake, check at different steps.

Exercise 6

1) \,x^{3}+ax^{2}-5x+b leaves remainder \,1 and \,7 when divided by \,x-2 and \,x+1 respectively. Find the values of \,a & \,b.

  • \,a=0, b=3
    • Let P\left(x\right)=x^{3}+ax^{2}-5x+b
    • \begin{array} {ll} \therefore P\left(2\right)=1 & P\left(-1\right)=7 \\ 8+4a-10+b=1 \qquad & -1+a+5+b=7\\ 4a+b=3\frac{\qquad}{}(1) & a+b=3 \frac{\qquad}{}(2)\\ \end{array}
    • (1)-(2) \; : \; 3a =0
    • \therefore a=0, b=3

2) \,ax^{3}-5x^{2}+bx+7 leaves remainder \,12 and \,-29 when divided by \,2x-1 and \,2x+3 respectively. Find the values of \,a & \,b.

  • \,a=2, b=12
    • Let P\left(x\right)=ax^{3}-5x^{2}+bx+7
    • \begin{array}{ll} \therefore P\left(\frac{1}{2}\right)=12 & P\left(-\frac{3}{2}\right)=-29 \\ \frac{1}{8}a-\frac{5}{4}+\frac{1}{2}b+7=12 \qquad & -\frac{27}{8}a-\frac{45}{4}-\frac{3}{2}b+7=-29\\ a-10+4b=40 & -27a-90-12b=-288\\ a+4b=50 \frac{\qquad}{}(1)& 27a+12b=198 \\ & 9a+4b=66\frac{\qquad}{}(2) \end{array}
    • (2)-(1)\;:\; 8a=16, \therefore a=2
    • 2+4b=50, \therefore b =12
    • \therefore a=2, b=12

3) \,3x^{3}-2x^{2}+ax+2 leaves a remainder \,b and \,-4b when divided by \,3x+1 and \,x+2 respectively. Find the values of \,a & \,b.

  • \,a=-7, b=4
    • Let P\left(x\right)=3x^{3}-2x^{2}+ax+2
    • \begin{array} {ll} \therefore P\left(-\frac{1}{3}\right)=b & P\left(-2\right)=-4b \\ -\frac{1}{9}-\frac{2}{9}-\frac{1}{3}a+2=b \qquad & -24-8-2a+2=-4b\\ -1-a+6=3b & 4b-2a= 30 \\ 3b+a =5\frac{\qquad}{}(1) & 2b-a=15 \frac{\qquad}{}(2)\\ \end{array}
    • (1)+(2)\;:\; 5b=20, \therefore b=4
    • 12+a=5, \therefore a=-7

4) \,x^{2}+ax-5 and \,ax^{2}-5x-a leaves the same remainder when divided by \,x+3. Find value of \,a

  • \,a=-1
    • Let P\left(x\right)=x^{2}+ax-5, Q\left(x\right)=ax^{2}-5x-a
    • P\left(-3\right)=Q\left(-3\right)
    • \,9-3a-5=9a+15-a
    • 11a=-11, \therefore a=-1

5) \,5x^{2}-2x+17 leaves the same remainder when divided by \,x+a and \,x-3a respectively. Find \,a, where a \neq 0

  • a=\frac{1}{5}
    • Let P\left(x\right)=5x^{2}-2x+17
    • P\left(-a\right)=P\left(3a\right)
    • \begin{array}{ll} 5a^{2}+2a+17 = 45a^{2}-6a+17 \\ 40a^{2}-8a =0 \\ a\left(5a-1\right) = 0 \\ a=0 \;\mbox{or}\; a=\frac{1}{5} \\ \therefore a=\frac{1}{5} \\ \end{array}

Quadratic Divisor

When P\left(x\right) is divided by a quadratic divisor, its remainder is a linear function.

Thus, to find its remainder without long division, we can't use remainder theorem directly. We will however, use the same steps as those in the proof of the theorem.

  • If P\left(x\right) is divided by \left(x-a\right)\left(x-b\right), its remainder is \,Ax+B
  • \therefore P\left(x\right)= \left(x-a\right)\left(x-b\right)Q\left(x\right)+Ax+B

Note : The most important step is writing the above equation.

Examples

  • Without using long division, find the remainder when \,x^{4}-x+1 is divided by \,x^{2}-1
  • Analysis : Divided by quadratic, not allowed to use long division
    • Let the remainder = \,Ax+B
    • \therefore \,x^{4}-x+1 = \left(x^{2}-1\right) Q\left(x\right) \,+Ax+B
    • Keep in mind we don't know (and don't need to find) Q\left(x\right)
    • Thus, we must choose value(s) of \,x such that \,x^{2}-1 is zero. Note that we need two values since we have two unknowns.
    • The values that we will put in are \,x=1 and \,x=-1
    • Remember that every \,x need to be substituted ( Q\left(x\right) don't need be written, though). Such as when \,x=1
      • LHS = \left(-1\right)^{4}-\left(-1\right)+1=1
      • RHS = \left(1-1\right)Q\left(1\right)+A\left(1\right)+B=A+B
    • \begin{align} & \mbox{When}\; x=1 \\ & 1=A+B \\ \end{align}
    • \begin{align} & \mbox{When}\; x=-1 \\ & 3=-A+B \end{align}
    • Next we solve the simultaneous equations
    • \begin{align} & \mbox{When}\; x=1 \\ & 1=A+B \frac{\qquad}{}(1)\\ & \mbox{When}\; x=-1 \\ & 3=-A+B \frac{\qquad}{}(2)\\ & (1)+(2)\;:\; 4 = 2B, \therefore B=2\\ & 1=A+2, \therefore A=-1\\ \end{align}
    • Finished? No. Have we answered the question? What did the question ask for?
    • Thus, the remainder is \,-x+2
    • Check: It's a bit too long to check by long division. Instead, here we check that we have calculated the two equations correctly, and have solve them correctly. Note that we can't put in any other values into the rearranged polynomial to check since we do not have Q\left(x\right).

  • f\left(x\right) leaves remainder \,1 and \,4 when divided by \,x-1 and \,x-2 respectively. Find the remainder when f\left(x\right) is divided by \left(x-1\right)\left(x-2\right)
  • Analysis : Note that we don't have f\left(x\right), we don't even know its degree, so how in the world are we going to calculate anything?
    • Let's just try to see what we can do with the given information
      • "f\left(x\right) leaves remainder \,1 and \,4 when divided by \,x-1 and \,x-2 respectively"
        • Thus, at least we know f\left(1\right)=1 and f\left(2\right)=4 (which doesn't seem of much use here since we don't have f\left(x\right), but never mind first)
        • We then look at the question. We need to find the remainder when f\left(x\right) is divided by \left(x-1\right)\left(x-2\right), which means that it is being divided by a quadratic divisor
        • So, at least we can write f\left(x\right)=\left(x-1\right)\left(x-2\right)Q\left(x\right)+Ax+B
        • We will then substitute values that makes \left(x-1\right)\left(x-2\right) zero. Which means we need f\left(1\right) and f\left(2\right)
        • But we don't have f\left(x\right) ! We need it to find f\left(1\right) and f\left(2\right). Or so we thought!

  • Solution:
    • f\left(1\right)=1, f\left(2\right)=4
    • f\left(x\right)=\left(x-1\right)\left(x-2\right)Q\left(x\right)+Ax+B
    • \mbox{When}\;x=1 \qquad \quad \mbox{When}\;x=2
    • f\left(1\right)=A+B \quad \quad f\left(2\right)=2A+B
    • 1=A+B\frac{\quad}{}(1) \quad 4=2A+B\frac{\quad}{}(2)
    • (2)-(1)\;:\; 3=A , \therefore B=-2
    • \therefore \mbox{Remainder} = 3x-2

Exercise 7

1) Without using long division, find the remainder of \,2x^{3}+x^{2}-16x-4 when divided by

  • a)\left(x-3\right)\left(2x+1\right) \,2x+5
      • 2x^{3}+x^{2}-16x-4= \left(x-3\right)\left(2x+1\right)Q\left(x\right)+Ax+B
      • \begin{align} & \mbox{When} \; x=3 \\ & 11=3A+B \frac{\qquad}{}(1)\\ & \mbox{When} \; x=-\frac{1}{2} \\ & 4=-\frac{1}{2}A+B \frac{\qquad}{}(2)\\ & (1)-(2)\;:\; 7 =\frac{7}{2}A \\ & \therefore A=2 \\ & 11=6+B, \therefore B=5\\ \end{align}
      • \therefore Remainder is \,2x+5

  • b)\,x^2-4 \,-8x
      • 2x^{3}+x^{2}-16x-4= \left(x^{2}-4\right)Q\left(x\right)+Ax+B
      • \begin{align} & \mbox{When} \; x=2 \\ & -16=2A+B \frac{\qquad}{}(1)\\ & \mbox{When} \; x=-2 \\ & 16=-2A+B \frac{\qquad}{}(2)\\ & (1)+(2)\;:\; 0 =2B \\ & \therefore B=0 \\ & -16=2A, \therefore A=-8\\ \end{align}
      • \therefore Remainder is \,-8x

2) \,ax^{3}+4x^{2}+bx-7 gives remainder \,13x-5 when divided by \left(x-1\right)\left(x+2\right). Find values of \,a and \,b.

  • \,a=3, b=8
    • \begin{align} & ax^{3}+4x^{2}+bx-7 = \left(x-1\right)\left(x+2\right)Q\left(x\right)+13x-5 \\ & \mbox{When}\; x=1 \\ & a+4+b-7=13-5 \\ & a+b = 11 \frac{\qquad}{}(1) \\ & \mbox{When}\; x=-2 \\ & -8a +16 -2b -7 = -26-5 \\ & -8a -2b = -40 \\ & 4a+b =20 \frac{\qquad}{}(2)\\ & (2)-(1) \;:\; 3a =9, \therefore a=3 \\ & 3+b=11, \therefore b=8 \\ &\therefore a=3, b=8 \end{align}

3) f\left(x\right) leaves remainder \,-21 and \,1 when divided by \left(x+5\right) and \left(2x-1\right) respectively. Find remainder when f\left(x\right) is divided by \,2x^{2}+9x-5

  • \,4x-1
    • \begin{align} & \left(x+5\right)\left(2x-1\right)=2x^{2}+9x-5 \\ & f\left(x\right)= \left(2x^{2}+9x-5\right)Q\left(x\right) +Ax+B \\ & \therefore f\left(x\right)= \left(x+5\right)\left(2x-1\right)Q\left(x\right) +Ax+B \\ \end{align}
    • \begin{array}{ll} f\left(-5\right)=-21 \qquad \qquad & f\left(\frac{1}{2}\right)=1 \\ -5A+B=-21\frac{\qquad}{}(1) & \frac{1}{2}A+B=1 \frac{\qquad}{}(2)\\ \end{array}
    • \begin{align} & (1)-(2) \;:\; -\frac{11}{2}A=-22, \therefore A=4 \\ & -20+B=-21, \therefore B=-1 \\ & \therefore \mbox{Remainder} \; = 4x-1 \end{align}
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