Polynomials Part3

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Contents

Factor Theorem

\,x-a is a factor of P\left(x\right) if and only if P\left(a\right)=0

Factors Of Polynomial

Before we go to the proof, let us ask ourselves, what is a factor of polynomial? Or more simply, what is a factor?

  • 6 =2\times 3 We thus say \,2 and \,3 are factors of \,6
  • 7 =2\times 3+1 We don't say \,2 and \,3 are factors of \,7

Thus, for polynomial

  • P\left(x\right)=\left(x-1\right)\left(x+2\right)\left(2x-3\right) \left(x-1\right),\;\left(x+2\right),\;\left(2x-3\right) are factors of P\left(x\right)
  • Q\left(x\right)=\left(x-1\right)\left(x+2\right)\left(2x-3\right)+2x+1 \left(x-1\right),\;\left(x+2\right),\;\left(2x-3\right) are not factors of Q\left(x\right)

What's the difference between the two? One doesn't have remainder, the other does.

If and only if

Whenever a theorem says if and only if, it means that the theorem works both ways. In this case, the theorem is actually a combination of

  • If \,x-a is a factor of P\left(x\right), then P\left(a\right)=0
  • If P\left(a\right)=0, then \,x-a is a factor of P\left(x\right)

Note that a lot of theorem (but not all) actually works both ways, even though the if and only if is not written explicitly.

However, if we are asked to prove a theorem which states if and only if, we are are required to prove it both ways.

Proof

  • Let's see what we know thus far
    • From example above, we can see that if there is a remainder, it wont be factor
    • In factor theorem, it says P\left(a\right)=0, but what does P\left(a\right) represents? Remainder
    • Obvious now, isn't it?

  • Let's write out the prove. We know we need to use Remainder Theorem, so we start from there, and then we also need to prove from both sides
    • When P\left(x\right) is divided by \,x-a, from Remainder Theorem, P\left(x\right)=\left(x-a\right)Q\left(x\right)+P\left(a\right)
      • If \,x-a is a factor P\left(x\right), thus P\left(a\right)=0
      • If P\left(a\right)=0, that is, P\left(x\right)=\left(x-a\right)Q\left(x\right), thus \,x-a is a factor of P\left(x\right)
    • \therefore \,x-a is a factor of P\left(x\right) if and only if P\left(a\right)=0

Usage

As with remainder theorem, we must learn how to use it properly.

To Prove Factor

Example : Show that \,x-1 is a factor of \,x^{2}-7x+6

  • Impossibly simple, right? No. Consider the following sample solutions(very common)
    • Solution 1 \,x=1, 1^{2}-7+6=0
    • Solution 2 \mbox{Let}\;P\left(x\right)=x^{2}-7x+6, P\left(1\right)=1^{2}-7+6=0
    • Which one is correct? First one? Second one? Both are wrong.

  • The same two things from remainder theorem apply here
    • Must name polynomial (if not given)
    • Must write conclusion

  • \begin{align}
& \mbox{Let}\;P\left(x\right)=x^{2}-7x+6 \\
& P\left(1\right)=1^{2}-7+6=0 \\
& \therefore x-1\;\mbox{is a factor of}\; P\left(x\right)
\end{align}

Polynomial vs Equation

Example :

P\left(x\right)=\left(x-1\right)\left(x+2\right)\left(2x-3\right) This is a polynomial

  • When P\left(x\right)=0 This is an equation (which involves the defined polynomial)
    • \therefore \left(x-1\right)\left(x+2\right)\left(2x-3\right) = 0
    • \therefore x=1, x=-2, x=\frac{3}{2} This is the solution to the equation
  • When P\left(x\right)=x+1 Another equation
    • \therefore \left(x-1\right)\left(x+2\right)\left(2x-3\right) = x+1

Zeroes, Factors, Roots

  • \,a is a zero of P\left(x\right) if P\left(a\right)=0 In other word, a zero is a value which will make the polynomial to have zero value
  • \,a is a root of an equation if is satisfies that equation (solution to equation)

Important Note

  • factors and zeroes of polynomial
  • roots of equation

Example

P\left(x\right)=\left(x-1\right)\left(x+2\right)\left(2x-3\right)

  • a) \,x-1, x+2, 2x-3 are factors of P\left(x\right)
  • b) 1,-2,-\frac{3}{2} are zeroes of P\left(x\right)
  • c) x=1,x=-2,x=-\frac{3}{2} are roots of the equation P\left(x\right)=0

Note that zeroes, factors and roots, though related, are different concepts. Giving one as the answer when asked for another will result in zero/less marks.

Relationship Between Zeroes, Factors, Roots

  • \,a is a zero of P\left(x\right)
  • \Leftrightarrow(if and only if)
  • \,x-a is a factor of P\left(x\right)
  • \Leftrightarrow
  • \,x=a is a root of P\left(x\right)=0

Writing Solutions

Take note that different type of questions will involve different ways of writing, even if we are dealing with the exact same numbers. Here we see some of the ways the questions can be asked for a relatively simple polynomial.

  • P\left(x\right)=2x^{3}-7x^{2}+2x+3
    • Show that \,x-1 is a factor of P\left(x\right)
      • P\left(1\right)=2-7+2+3=0
      • \therefore x-1 \mbox{ is a factor of } P\left(x\right)
    • Show that \,1 is a zero of P\left(x\right)
      • P\left(1\right)=2-7+2+3=0
      • \therefore 1 \mbox{ is a zero of } P\left(x\right)
    • Show that \,x=1 is a root of P\left(x\right)=0
      • \begin{align}
& P\left(x\right)=0 \\
& \therefore 2x^{3}-7x^{2}+2x+3 =0 \\
\end{align}
      • \begin{align}
\mbox{When } x & =1, \\
\mbox{R.H.S. }& =2-7+2+3 \\
& = 0 \\
& = \mbox{L.H.S.}
\end{align}
      • \therefore x=1 \mbox{ is a root of } P\left(x\right)=0
  • Hence (question continue from above)
    • Factorize P\left(x\right) completely
      • x-1 \mbox{ is a factor of } P\left(x\right)
      • \therefore P\left(x\right)=\left(x-1\right)\left(2x^{2}-5x-3\right)
      • =\left(x-1\right)\left(2x+1\right)\left(x-3\right)
    • Find all factors of P\left(x\right)
      • \,x-1,2x+1,x-3
    • Find the other factors of P\left(x\right)
      • \,\mbox{Other factors are } 2x+1 \mbox{and } x-3
    • Find all roots of the equation P\left(x\right)=0
      • \, x=1, x=-\frac{1}{2}, x=3
    • Solve the equation P\left(x\right)=0
      • \, x=1, x=-\frac{1}{2}, x=3

Exercise 8

1) Show that \,x-a is a factor of 2x^{2}-3\left(a-1\right)x+a\left(a-3\right)

    • \mbox{Let}\; P\left(x\right)=2x^{2}-3\left(a-1\right)x+a\left(a-3\right)
    • \begin{align}
P\left(a\right)& =2a^{2}-3\left(a-1\right)a+a\left(a-3\right)\\
& =2a^{2}-3a^{2}+3a+a^{2}-3a \\
& = 0
\end{align}
    • \therefore x-a \; \mbox{is a factor of}\;2x^{2}-3\left(a-1\right)x+a\left(a-3\right)

2) Show that \,x+3 is factor of P\left(x\right)=x^{3}-x^{2}-17x-15. Hence, solve the equation P\left(x\right)=0

  • \,x=-3, x=-1, x=5
    • \begin{align}
& P\left(-3\right)=-27-9+51-15=0 \\
& \therefore x+3 \mbox{ is a factor of } P\left(x\right)\\
\end{align}
    • \begin{align}
\therefore P\left(x\right) & = \left(x+3\right)\left(x^{2}-4x-5\right) \\
& = \left(x+3\right)\left(x+1\right)\left(x-5\right)
\end{align}
    • \begin{align}
& \mbox{If}\;P\left(x\right) = 0 \\
& \left(x+3\right)\left(x+1\right)\left(x-5\right) =0 \\
& \therefore x=-3, x=-1, x=5
\end{align}

3) Show that -\frac{1}{2} is a zero of P\left(x\right)=2x^{3}-x^{2}-5x-2. Hence, factorize the polynomial completely.

  • P\left(x\right)=\left(2x+1\right)\left(x-2\right)\left(x+1\right)
    • \begin{align}
P\left(-\frac{1}{2}\right)& =2\left(-\frac{1}{8}\right)-\frac{1}{4}+\frac{5}{2}-2
& =0
\end{align}
    • \therefore -\frac{1}{2} \mbox{ is a zero of } P\left(x\right)
    • \therefore 2x+1 \mbox{ is a factor of } P\left(x\right)
    • \begin{align}
\therefore P\left(x\right)& = \left(2x+1\right)\left(x^{2}-x-2\right) \\
& = 
\left(2x+1\right)\left(x-2\right)\left(x+1\right)
\end{align}

4) Show that \,-2 is a root to the equation \,3x^{3}-7x^{2}-22x+8=0. Hence, find the other roots.

  • x=4, x=\frac{1}{3}
    • \, \mbox{When } x=-2
    • \begin{align}
\mbox{R.H.S } & = -24-28+44+8 \\
& = 0 \\
& = \mbox{L.H.S}
\end{align}
    • \therefore -2 \mbox{ is a root to the equation.}
    • \begin{align}
& 3x^{3}-7x^{2}-22x+8=0 \\
& \left(x+2\right)\left(3x^{2}-13x+4\right)=0\\
& \left(x+2\right)\left(3x-1\right)\left(x-4\right)=0\\
& \therefore x=-2, x=\frac{1}{3}, x=4 \\
& \therefore \mbox{ the other roots are } x=4, x=\frac{1}{3}
\end{align}

Interpreting Information into Equations

Example :

\,x-2 is a factor of P\left(x\right)

Thus, P\left(2\right)=0

  • Other forms : can be divided / can be divided exactly / divisible

Exercise 9

1) Complete the following table

Information Equation
\,x-3 is a factor of P\left(x\right) P\left(3\right)=0
P\left(x\right) have factors \,x-2 and \,2x+1 P\left(2\right)=0, P\left(-\frac{1}{2}\right)=0
P\left(x\right) is divisible by \left(x-a\right)\left(x-b\right) P\left(a\right)=0, P\left(b\right)=0
P\left(x\right) and G\left(x\right) have a common factor P\left(a\right)=G\left(a\right)=0(Assume common factor \,x-a)
P\left(x\right) can be divided by \,x-1 but leaves remainder \,2 when divided by \,3x+2 P\left(1\right)=0, P\left(-\frac{2}{3}\right)=2

2) The polynomial \,ax^{3}-5x^{2}+bx+7 has a factor \,2x+1 and leaves remainder \,25 when divided by \,x-2. Find the values of \,a and \,b.

  • \,a=2,b=11
    • \mbox{Let}\;P\left(x\right)=ax^{3}-5x^{2}+bx+7
    • \begin{array}{ll}
P\left(-\frac{1}{2}\right)=0 & P\left(2\right)=25 \\
-\frac{1}{8}a-\frac{5}{4}-\frac{1}{2}b+7=0 & 8a-20+2b+7=25 \\
-a-10-4b+56=0 & 8a+2b=38 \\
a+4b=46 \frac{\qquad}{}(1) & 16a+4b=76 \frac{\qquad}{}(2)
\end{array}
    • (2)-(1):\; 15a=30 \therefore a=2
    • 2+4b=46 \therefore b=11
    • \therefore a=2, b=11

3) P\left(x\right)=2x^{3}+ax^{2}+bx+30 is divisible by \,x+3 and \,x-2. Find the values of \,a and \,b.

  • \,a=-3, b=-17
    • \begin{array}{ll}
P\left(-3\right)=0 & P\left(2\right)=0 \\
-54+9a-3b+30=0 & 16+4a+2b+30=0 \\
9a-3b=24 & 4a+2b =-46 \\
3a-b=8\frac{\quad}{}(1) & 2a+b=-23\frac{\quad}{}(2) \\
\end{array}
    • (1)+(2):\; 5a=-15 \therefore a=-3
    • -9-b=8 \therefore b=-17
    • \therefore a=-3, b=-17

4) P\left(x\right)=2x^{2}+3x+c and G\left(x\right)=x^{2}-2x+c (where c \neq 0). Find the value of \,c and that common factor.

  • \,c=-35, x+5
    • \,\mbox{Let common factor} = x-a
    • \begin{align}
& P\left(a\right)=G\left(a\right)=0 \\
& 2a^{2}+3a+c=a^{2}-2a+c \\
& a^{2}+5a=0 \\
& a\left(a+5\right)=0 \\
& a=0, a=-5 \\
& \mbox{When}\; a=0,\; c=0 \mbox{(rejected)} \\
& \mbox{When}\; a=-5,\; 25+10+c=0 \therefore c=-35 \\
& \therefore c=-35, \mbox{Common factor}=x+5 \\
\end{align}

Factorizing Polynomial

To fully factorize a cubic (or higher) polynomial, we need to obtain at least one factor first by trial-and-error method.

That is, we need to find a value \,a such that P\left(a\right)=0, \therefore x-a is a factor.

Trial and Error

Try factors of the constant

  • P\left(x\right)=\ldots +6 try \pm 1, \pm 2, \pm 3 , \pm 6
  • P\left(x\right)=\ldots -5 try \pm 1, \pm 5

Examples

  • Factorize \,6x^{3}+11x^{2}-3x-2
  • Take note that we are expected in this case to show full working, thus use the calculator to check, not do.
    • We should be using factor theorem, so firstly we..
      • \mbox{Let }P\left(x\right)=6x^{3}+11x^{2}-3x-2
    • Then, we will need to use trial and error to find (at least ) one factor
      • Since the constant is \,-2, we will thus try \,1,-1,2,-2 until we get one. We might need to count a few before we get it, so might as well enter the formula to count quickly (and precisely).
        • \mbox{Try }P\left(1\right) \,=12
        • P\left(-1\right) \,=6
        • P\left(2\right) \,=84
        • P\left(-2\right) \,=0
        • \therefore x+2 \mbox{ is a factor}
          • DO NOT cut away the first three just because they are not factors, it's part of the process. In fact, if you directly write P\left(-2\right)=0 without the first three, it simply implies that you are cheating...
    • We can then use long division/compare coefficients to find the other factors
      • \begin{align}
\therefore P\left(x\right)& =\left(x+2\right)\left(6x^{2}-x-1\right)\\
& =\left(x+2\right)\left(3x+1\right)\left(2x-1\right)
\end{align} or \begin{align}
\therefore P\left(x\right)& =\left(x+2\right)\left(6x^{2}+Bx-1\right)\\
& \mbox{Coefficient of } x^{2} : 11=12+B \\
& \therefore B=-1 \\
\therefore P\left(x\right)& =\left(x+2\right)\left(6x^{2}-x-1\right)\\
& =\left(x+2\right)\left(3x+1\right)\left(2x-1\right)
\end{align}

  • Factorize P\left(x\right)=x^{4}+x^{3}-15x^{2}+23x-10
  • Note that if we only find one factor through trial and error, we will have to do long division, which gives us a cubic, which we then need to do trial and error again. So in this case, we try until we get two factors.
    • \begin{align}
\mbox{Try } & P\left(1\right)=0 \\
& P\left(-1\right)=48 \\
& P\left(2\right)=0 \\
\end{align}
    • \therefore x-1 \mbox{ and } x-2 \mbox { are factors }
    • \left(x-1\right)\left(x-2\right)=x^{2}-3x+2
    • \begin{array}{rl}
& \qquad \qquad \quad \;\; x^{2}+\;4x-\;\;5 \\
& x^{2}-3x+2 \overline{)\;x^{4}+\;x^{3}-15x^{2}+23x-10} \\
& \qquad \qquad \;\; -\underline{\left(x^{4}-3x^{3}+\;2x^{2}\right)} \\
& \qquad \qquad \qquad \quad \;\; 4x^{3}-17x^{2}+23x \\
& \qquad \qquad \qquad \;-\underline{\left(4x^{3}-\;12x^{2}+\;8x \right)} \\
& \qquad \qquad \qquad \qquad \quad\;\;\; -5x^{2} +\;15x-10 \\
& \qquad \qquad \qquad \qquad \;\; -\underline{\left(-5x^{2} +\;15x-10\right)} \\
& \qquad \qquad \qquad \qquad \qquad \qquad \qquad - \\ \end{array}
    • \therefore P\left(x\right)
      • =\left(x-1\right)\left(x-2\right)\left(x^{2}+4x-5\right)
      • =\left(x-1\right)\left(x-2\right)\left(x-1\right)\left(x+5\right)
      • =\left(x-1\right)^{2}\left(x-2\right)\left(x+5\right)

Note

1) Trial-and-error is useful to find one or two factors, so that the rest can be found by usual methods. We don't however, use trial-and-error to find ALL factors. This is because there are often factors that can't be found by this method

  • non integer factors such as \,2x+1 (actually it's possible, but very tedious to try out fractions)
  • repeated factors (this is also one of the reason that calculator is used to check only, not do)

2) Quadratic polynomials that have complex roots generally does not need to be factored further (since if will involve complex factors)

  • P\left(x \right) =\left(x-1\right)\underbrace{\left(x^{2}+2x+5 \right)}_{complex\;roots}
  • P\left(x \right) =\left(x+1\right)\underbrace{\left(2x^{2}-6x+15 \right)}_{complex\;roots}

Exercise 10

Factorize the following polynomial completely

1) \,2x^{3}-5x^{2}-4x+3

  • \left(x+1\right)\left(x-3\right)\left(2x-1\right)
    • \begin{align}
& \mbox{Let }P\left(x\right)= 2x^{3}-5x^{2}-4x+3 \\
& \mbox{Try }P\left(1\right)=-4 \\
& \qquad P\left(-1\right)=0 \\
& \therefore x+1 \mbox{ is a factor } \\
& \therefore P\left(x\right)= \left(x+1\right)\left(2x^{2}-7x+3\right) \\
& \therefore P\left(x\right)= \left(x+1\right)\left(x-3\right)\left(2x-1\right)
\end{align}

2) \,3x^{3}-16x^{2}+6x-5

  • \left(x-5\right)\left(3x^{2}-x+1\right)
    • \begin{align}
& \mbox{Let }P\left(x\right)= 3x^{3}-16x^{2}+6x-5 \\
& \mbox{Try }P\left(1\right)=-12 \\
& \qquad P\left(-1\right)=-30 \\
& \qquad P\left(5\right)=0 \\
& \therefore x-5 \mbox{ is a factor } \\
& \therefore P\left(x\right)= \left(x-5\right)\left(3x^{2}-x+1\right) \\
\end{align}

3) \,x^{4}+3x^{3}-5x^{2}-3x+4

  • \left(x-1\right)^{2}\left(x+1\right)\left(x+4\right)
    • \begin{align}
& \mbox{Let }P\left(x\right)= \,x^{4}+3x^{3}-5x^{2}-3x+4 \\
& \mbox{Try }P\left(1\right)=0 \\
& \qquad P\left(-1\right)=0 \\
& \therefore x+1 \mbox{ and } x-1 \mbox{ are factors } \\
\end{align}
    •  \begin{array}{rl} 
& \qquad \quad \;\; x^{2}+3x-4 \\
& x^{2}-1\overline{)\;x^{4}+3x^{3}-5x^{2}-3x+4} \\ 
& \qquad -\underline{\left(x^{4}\qquad\quad-x^{2}\right)} \\ 
& \qquad \qquad \quad \;\; 3x^{3}-4x^{2}-3x \\
& \qquad \qquad \;-\underline{\left(3x^{3}\qquad\quad-3x \right)} \\
& \qquad \qquad \qquad \quad \; -4x^{2}\qquad\quad+4 \\
& \qquad \qquad \qquad -\underline{\left(-4x^{2}\qquad\quad+4\right)} \\
& \qquad \qquad \qquad \qquad \qquad \quad - \\
\end{array}
    • \begin{align}
\therefore P\left(x\right) & =\left(x+1\right)\left(x-1\right)\left(x^{2}+3x-4\right)\\
& =\left(x+1\right)\left(x-1\right)\left(x+4\right)\left(x-1\right)\\
& =\left(x-1\right)^{2}\left(x-1\right)\left(x+4\right)\\
\end{align}

4) \,2x^{4}+x^{3}+7x-10

  • \left(x-1\right)\left(x+2\right)\left(2x^{2}-x+5\right)
    • \begin{align}
& \mbox{Let }P\left(x\right)= \,2x^{4}+x^{3}+7x-10 \\
& \mbox{Try }P\left(1\right)=0 \\
& \qquad P\left(-1\right)=-16 \\
& \qquad P\left(2\right)=44 \\
& \qquad P\left(-2\right)=0 \\
& \therefore x-1 \mbox{ and } x+2 \mbox{ are factors } \\
& \left(x-1\right)\left(x+2\right)= x^{2}+x-2 
\end{align}
    •  \begin{array}{rl}
& \qquad \qquad \quad \; 2x^{2}-\;x\;\;\;+5 \\
& x^{2}+x-2 \overline{)\;2x^{4}+\;x^{3}\;-0x^{2}+7x-10} \\ 
& \qquad \qquad -\underline{\left(2x^{4}+2x^{3}-4x^{2}\right)} \\
& \qquad \qquad \qquad \quad \;\;\; -x^{3}+4x^{2}+7x \\
& \qquad \qquad \qquad \;\; -\underline{\left(-x^{3}-\;x^{2}+\;2x \right)} \\
& \qquad \qquad \qquad \qquad \qquad\;\; 5x^{2} +\;5x-10 \\
& \qquad \qquad \qquad \qquad \quad\; -\underline{\left(5x^{2} +\;5x-10\right)} \\
& \qquad \qquad \qquad \qquad \qquad \qquad \qquad - \\
\end{array}
    • \therefore P\left(x\right)=\left(x-1\right)\left(x+2\right)\left(2x^{2}-x+5\right)

Long Questions

A few important things to keep in mind when doing multi-part questions

  • Always make sure first part is correct (especially when it involves finding value of constants that defines the polynomial) BEFORE continuing to subsequent parts
  • Quite often, the first factor(s) are already given in the first part of the question. We can use them directly when in the subsequent parts .
  • Always give answer in the form asked by the questions ; factorize completely, find all factors, find other factors, find all roots, etc.

Hence/Thus

Whenever we see the word Hence or Thus in the question, it means

  • We MUST use the information we obtained from the EARLIER part of the question to solve the NEXT part
  • Failure to do so will result in ZERO marks, EVEN if you get the CORRECT answers
  • This applies to ALL types of questions from any topics
  • For some particular types of questions (refer to matrices), or when the connection between the different parts of the questions is too obvious, sometimes the word hence/thus is not written, but still we are expected to use that connection (otherwise, zero marks)

Hence, or Otherwise

If however, we see Hence, or otherwise, it means that we MAY use other methods. However, it is almost guaranteed that those other methods will be more time consuming, more susceptible to careless mistakes, and well, less fun than the hence method.

Example

Factorize P\left(x\right)=x^{3}-x completely. Hence, solve P\left(x\right)=6x-6

  • Analyze
    • The question asks us to factorize, and then hence, solve the given equation.
    • Remember that hence implies we must use information obtained from earlier part to solve. In this case,
      • First part we will get the factored from of P\left(x\right)=x^{3}-x
      • Second part we have to solve the equation, and we must use information of the first part, which is the factors/ factored form of P\left(x\right)
  • P\left(x\right)=x^{3}-x
    • =x\left(x^{2}-1\right)
    • =x\left(x-1\right)\left(x+1\right)

  • P\left(x\right)=6x-6
    • \therefore x\left(x-1\right)\left(x+1\right)=6x-6
    • This is where it can be tricky. Remember that we are required to use the factors already found /factored form
      • Thus, if we go and do \begin{align}
\therefore & x^{3}-x=6x-6 \\
& x^{3}-7x+6=0 \\
\end{align}, we will be given zero marks.
      • Instead, since the LHS is already factored, is there anything we can do to the RHS?
      • \,6x-6 =6\left(x-1\right)
      • That is, x{\color{Blue}\left(x-1\right)}\left(x+1\right)=6{\color{Blue}\left(x-1\right)}
    • x\left(x-1\right)\left(x+1\right)=6\left(x-1\right)
    • \left(x-1\right)\left[x\left(x+1\right)-6\right]=0
    • \left(x-1\right)\left(x^{2}+x-6\right)=0
    • \left(x-1\right)\left(x+3\right)\left(x-2\right)=0
    • \,x=1,x=2,x=-3

Exercise 11

1) P\left(x\right)=ax^{3}+5x^{2}+bx-28 has a factor \,x-2 and leaves remainder \,-10 when divided by \,x+3. Find the values of \,a and \,b. Factorize the polynomial completely.

  • a=1, b=0; P\left(x\right)=\left(x-2\right)\left(x^{2}+7x+14\right)
    • \begin{array}{ll}
P\left(2\right)=0 & P\left(-3\right)=-10 \\
8a+20+2b-28=0 \qquad & -27a+45-3b-28=-10 \\
8a+2b=8 & 27a+3b = 27 \\
4a+b=4\frac{\quad}{}(1) & 3a+b=3 \frac{\quad}{}(2) \\
\end{array}
    • \,(2)-(1):\;a=1, b=0
    • \begin{align}
\therefore P\left(x\right)& = x^{3}+5x^{2}-28 \\
& =\left(x-2\right)\left(x^{2}+7x+14\right) \\
\end{align}

2) P\left(x\right)=x^{4}-2x^{3}+ax^{2}+bx+36 is divisible by \,x-3 and \,x+2. Find the values of \,a and \,b. Factorize the polynomial completely.

  • P\left(x\right)=\left(x-3\right)^{2}\left(x+2\right)^{2}
    • \begin{array}{ll}
P\left(3\right) = 0 & P\left(-2\right) = 0 \\
81-54+9a+3b+36 = 0 \qquad & 16+16+4a-2b+36 =0 \\
9a+3b=-63 & 4a-2b=-68 \\
3a+b=-21\frac{\quad}{}(1) & 2a-b=-34\frac{\quad}{}(2) \\\\
\end{array}
    • \begin{align}
&(2)+(1):\;5a=-55, \therefore a =-11 \\
& \therefore -22-b=-34, \therefore b=12 \\
& \therefore a=-11, b=12
\end{align}
    • \therefore P\left(x\right)=x^{4}-2x^{3}-11x^{2}+12x+36
    • \left(x-2\right)\left(x+3\right)=x^{2}-x-6
    • \begin{array}{rl}
& \qquad \qquad \quad \; x^{2}-\;\;x-\;\;6 \\
& x^{2}-x-6 \overline{)\;x^{4}-2x^{3}-11x^{2}+12x+36} \\
& \qquad \qquad -\underline{\left(x^{4}-\;\;x^{3}-\;6x^{2}\right)} \\
& \qquad \qquad \qquad \quad -x^{3}-\;\;5x^{2}+12x \\
& \qquad \qquad \qquad -\underline{\left(-x^{3}+\;\;\;x^{2}+\;6x \right)} \\
& \qquad \qquad \qquad \qquad \quad\;\;\; -6x^{2} +\;6x+\;36 \\
& \qquad \qquad \qquad \qquad \;\; -\underline{\left(-6x^{2} +\;6x+\;36\right)} \\
& \qquad \qquad \qquad \qquad \qquad \qquad \qquad - \\ 
\end{array}
    • \begin{align}
\therefore P\left(x\right) & = \left(x-2\right)\left(x+3\right)\left(x^{2}-x-6\right) \\
& =\left(x-2\right)\left(x+3\right)\left(x-2\right)\left(x+3\right) \\
& =\left(x-2\right)^{2}\left(x+3\right)^{2}
\end{align}

3) Show that \,2x-1 is a factor of P\left(x\right)=2x^{3}-13x^{2}+26x-10. Hence, solve the equation P\left(x\right)=6x^{2}-23x+10

  • x=\frac{1}{2},x=4,x=5
    • \begin{align}
P\left(\frac{1}{2}\right) & = \frac{1}{4}-\frac{13}{4}+13-10 \\
& = 0\\
\end{align}
    • \begin{align}
& \therefore 2x-1 \mbox{ is a factor of } P\left(x\right) \\
& \therefore P\left(x\right) = \left(2x-1\right)\left(x^{2}-6x+10\right) \\
& \mbox{Given}\;  P\left(x\right) = 6x^{2}-23x+10 \\
& \therefore \left(2x-1\right)\left(x^{2}-6x+10\right) = 6x^{2}-23x+10 \\
& \left(2x-1\right)\left(x^{2}-6x+10\right) = \left(2x-1\right)\left(3x-10\right) \\
& \left(2x-1\right)\left(x^{2}-6x+10\right) - \left(2x-1\right)\left(3x-10\right)=0 \\
& \left(2x-1\right)\left[\left(x^{2}-6x+10\right) - \left(3x-10\right)\right]=0 \\
& \left(2x-1\right)\left(x^{2}-9x+20\right)=0 \\
& \left(2x-1\right)\left(x-4\right)\left(x-5\right)=0 \\
& x=\frac{1}{2}, x=4, x=5 \\
\end{align}

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