\(f(x)\) is a trinomial.
True
5.1 Revision
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End of chapter exercises
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5.2 Cubic polynomials
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Chapter 5: Polynomials
- Proofs of theorems are not examinable.
- Be careful of spending too much time on revision. Learners should be familiar with these factorisation methods from Grade 10 and 11.
- Explain terminology carefully. It is very important for learners to understand what a factor is and how it relates to the graphs of quadratic and cubic functions (covered in Differential Calculus chapter).
- Long division and synthetic division are included as an introduction to the remainder theorem. Do not spend too much time on these techniques.
- It is important for learners to understand that a remainder of zero means that divisor is a factor.
- Encourage learners to practise factorising by inspection.
5.1 Revision (EMCGQ)
Identifying polynomials (EMCGR)
| Terminology: | |
| Polynomial |
An expression that involves one or more variables having different powers and coefficients. \(a_{n}x^{n} + \ldots + a_2x^{2} + a_{1}x + a_{0}, \text{ where } n \in \mathbb{N}_0\) |
| Monomial |
A polynomial with one term. For example, \(7a^{2}b \text{ or } 15xyz^{2}\). |
| Binomial |
A polynomial that has two terms. For example, \(2x + 5z \text{ or } 26 - g^{2}k\). |
| Trinomial |
A polynomial that has three terms. For example, \(a - b + c \text{ or } 4x^2 + 17xy - y^3\). |
| Degree/Order |
The degree, also called the order, of a univariate polynomial is the value of the highest exponent in the polynomial. For example, \(7p - 12p^2 + 3p^5 + 8\) has a degree of \(\text{5}\). |
It is important to notice that the definition of a polynomial states that all exponents of the variables must be elements of the set of natural numbers. If an expression contains terms with exponents that are not natural numbers, then it is not a polynomial.
The following examples are not polynomials:
\begin{align*} & \frac{3}{y} - 4y^{2} + 1\\ & 5 \sqrt{k} + k - 2k^{2} \\ & x^{-2} + 3 + 7x^{2} \\ & t^{2} - 4t + 6t^{\frac{1}{3}} \end{align*}More on polynomials
Discuss whether the following statements are true or false:
- The expression \(3y^2 + 2y - 4\) is a trinomial of degree \(\text{2}\).
- \(25z^{5} - 36\sqrt{z}\) is a binomial of order \(\text{5}\).
- \(\text{25}\) is a constant polynomial of degree \(\text{0}\).
- \(3x^{2} - 2x - 5\) is a quadratic polynomial.
- The expression \(23b^{-\text{2}}\) is a monomial because it only has one term.
- \(\text{0}\) is a constant polynomial of undefined degree.
- A cubic polynomial has three terms and all the exponents are natural numbers.
- \begin{align*} \text{Given the expression } &: \frac{1}{t} - 3t^{2} + 1 \\ \text{If we multiply by } t &: 1 - 3t^{3} + t, \\ \text{we get a trinomial} &\text{ of degree } 3 \end{align*}
Identifying polynomials
Textbook Exercise 5.1
Given \(f(x) = 2x^{3} + 3x^{2} - 1\), determine whether the following statements are true or false. If false, provide the correct statement.
The coefficient of the \(x\) is zero.
True
\(f \left( \frac{1}{2} \right) = \frac{1}{12}\).
\begin{align*}
f(x) &= 2x^{3} + 3x^{2} - 1 \\
f \left( \frac{1}{2} \right) &= 2\left( \frac{1}{2} \right)^{3} + 3\left( \frac{1}{2} \right)^{2} -
1 \\
&= 2\left( \frac{1}{8} \right) + 3\left( \frac{1}{4} \right)- 1 \\
&= \frac{1}{4} + \frac{3}{4} - 1 \\
&= 1 - 1 \\
&= 0 \\
\therefore & \text{ False}
\end{align*}
\(f(x)\) is of degree \(\text{3}\).
True
The constant term is \(\text{1}\).
False: \(-\text{1}\)
\(f(x)\) will have \(\text{3}\) real roots.
True
Given \(g(x) = 2x^{3} - 9x^{2} + 7x + 6\), determine the following:
the number of terms in \(g(x)\).
\(\text{4}\)
the degree of \(g(x)\).
\(\text{3}\)
the coefficient of the \(x^{2}\) term.
\(-\text{9}\)
the constant term.
\(\text{6}\)
Determine which of the following expressions are polynomials and which are not. For those that are not polynomials, give reasons.
\(y^{3} + \sqrt{5}\)
Cubic polynomial
\(-x^{2} - x - 1\)
Quadratic polynomial
\(4\sqrt{k} - 9\)
Not a polynomial; in \(k^{\frac{1}{2}}\) the exponent is not a natural number.
\(\frac{2}{p} + p + 3\)
Not a polynomial; in \(p^{-1}\) the exponent is not a natural number.
\(x(x - 1)(x - 2) - 2\)
Cubic polynomial
\(\left( \sqrt{m} - 1 \right) \left( \sqrt{m} + 1 \right)\)
\begin{align*}
\left( \sqrt{m} - 1 \right) \left( \sqrt{m} + 1 \right) &= \left( \sqrt{m} \right)^{2} + \sqrt{m} -
\sqrt{m} - 1 \\
&= m - 1
\end{align*}
Linear polynomial
\(t^{0} - 1\)
\begin{align*}
t^{0} - 1 &= 1 - 1 \\
&= 0
\end{align*}
Zero polynomial
\(16y^{7}\)
Polynomial; degree \(\text{7}\)
\(- \frac{x^{3}}{2} + 5x^{2} + \frac{x}{3} - 11\)
Cubic polynomial
\(4b^{0} + 3b^{-1} + 5b^{2} - b^{3}\)
Not a polynomial; in \(b^{-1}\) the exponent is not a natural number.
Peter's mathematics homework is shown below. Find and correct his mistakes.
Homework:
Given \(p(x) = x + \dfrac{4}{x} - 5\), answer the following questions:
- Simplify the expression.
- Is \(p(x)\) a polynomial?
- What is the coefficient of the \(x\) term?
Peter's answers:
- \[\begin{array}{rll} p(x) &= x + \frac{4}{x} - 5 & (\text{restrictions: } x \ne 0) \\ &= x^{2} + 4 - 5x & (\text{multiply through by } x) \\ &= x^{2} - 5x + 4 & (\text{write in descending order}) \\ &= (x - 1)(x + 4) & (\text{factorise, quadratic has two roots}) \end{array}\]
- Yes, because it can be simplified to have exponents that are all natural numbers. It is a quadratic binomial because the highest exponent is two and there are only two terms; \((x-1)\) and \((x+4)\).
- Before I simplified, the coefficient of the \(x\) term was nothing and after I simplified it became \(\text{5}\).
Peter made the following mistakes:
-
This is not an equation, therefore Peter must not multiply through by \(x\). Peter did not factorise correctly, he made an error with the sign in the second bracket
. \[\begin{array}{rll} p(x) &= x + \frac{4}{x} - 5 & (\text{restrictions: } x \ne 0) \\ &= \frac{x^{2} + 4 - 5x}{x} & (\text{common denominator is } x) \\ &= \frac{x^{2} - 5x + 4}{x} & (\text{write in standard form}) \\ &= \frac{(x - 1)(x - 4)}{x} & (\text{factorise and be careful of the signs}) \end{array}\] - The original expression is not a polynomial because the exponent of the second term \(\left( \frac{4}{x} = 4x^{-1}\right)\) is not a natural number. Peter's simplified expression is a quadratic trinomial because it has three terms and two factors.
- In the original expression, the coefficient of the \(x\) term is \(\text{1}\). After Peter wrongly simplified the expression, the coefficient of the \(x\) term was \(-\text{5}\) and not \(\text{+5}\). Always remember that the coefficient includes the sign and the number in front of the variable.
Quadratic polynomials (EMCGS)
In earlier grades we learnt the following useful techniques and methods for factorising an expression:
- taking out a common factor
- factorising the difference of two squares
- grouping in pairs
- factorising the sum and difference of two cubes
We also looked at the different methods for factorising quadratic expressions:
- factorising by inspection
- completing the square
- using the quadratic formula
- making a suitable substitution
It is important to revise these methods; we use the quadratic formula for factorising cubic polynomials and we also use completing the square to find the equation of a circle in Chapter 7.
| Terminology: | |
| Variable | A symbol used to represent an unknown numerical value. For example, \(a, b, x, y,\alpha, \theta\). |
| Coefficient | The number or parameter that is multiplied by the variable of an expression. |
| Expression | A term or group of terms consisting of numbers, variables and the basic operators (\(+, -, \times, \div\)). |
| Univariate expression | An expression containing only one variable. |
| Equation | A mathematical statement that asserts that two expressions are equal. |
| Identity | A mathematical relationship that equates one expression with another. |
| Solution | A value or set of values that satisfies the original problem statement. |
| Root/Zero | A root, also referred to as the “zero”, of an equation is the value of \(x\) such that \(f(x)=0\) is satisfied. |
Worked example 1: Solving quadratic equations using factorisation
Solve for \(x\): \[\frac{3x}{x+2} + 1 = \frac{4}{x+1}\]
Determine the restrictions
The restrictions are the values for \(x\) that would result in the denominator being equal to \(\text{0}\), which would make the fraction undefined. Therefore \(x ≠ -\text{2}\) and \(x ≠ -\text{1}\).
Determine the lowest common denominator
The lowest common denominator is \(\left(x+2\right)\left(x+1\right)\).
Multiply each term in the equation by the lowest common denominator and simplify
\begin{align*} \frac{3x(x+2)(x+1)}{x+2} + (x+2)(x+1) &= \frac{4(x+2)(x+1)}{x+1} \\ 3x(x+1) + (x+2)(x+1) &= 4(x+2) \\ 3x^2 + 3x + x^2+3x+2 &= 4x+8 \\ 4x^2 + 2x - 6 &= 0 \\ 2x^2 + x - 3 &= 0 \end{align*}Factorise and solve the equation
\begin{align*} (2x + 3)(x - 1)&= 0\\ 2x + 3 =0 & \text{ or } x-1 = 0 \\ x = -\frac{3}{2} & \text{ or } x = 1 \end{align*}Check the solution by substituting both answers back into the original equation
Write the final answer
Therefore \(x=-1\frac{1}{2}\) or \(x = 1\).
Worked example 2: Using the quadratic formula
Find the roots of the function \(f(x)= 3x^2 + 4x - 8\).
Finding the roots
To determine the roots of \(f(x)\), we let \(3x^2 + 4x - 8 = 0\).
Check whether the expression can be factorised
The expression cannot be factorised, so the general quadratic formula must be used.
Identify the coefficients to substitute into the formula
\[a = 3; \qquad b = 4; \qquad c = -8\]
Apply the quadratic formula
\begin{align*} x &= \frac{-b±\sqrt{{b}^{2}-4ac}}{2a} \\ &= \frac{-\left(4\right)±\sqrt{{\left(4\right)}^{2}-4\left(3\right)\left(-8\right)}}{2\left(3\right)} \\ &= \frac{-4±\sqrt{16 + 96}}{6} \\ &= \frac{-4±\sqrt{112}}{6} \\ &= \frac{-4±\sqrt{16 \times 7}}{6} \\ &= \frac{-4±4\sqrt{7}}{6} \\ &= \frac{-2±2\sqrt{7}}{3} \end{align*}Write the final answer
Therefore \(x = \dfrac{-2+2\sqrt{7}}{3}\) or \(x = \dfrac{-2-2\sqrt{7}}{3}\)
Worked example 3: Solving quadratic equations by completing the square
Solve by completing the square: \(y^2-10y-11=0\)
The equation is already in the form \(ax^2 + bx + c = 0\)
Make sure the coefficient of the \(y^2\) term is equal to \(\text{1}\)
\[y^2-10y-11=0\]Take half the coefficient of the \(y\) term and square it, then add and subtract it from the equation
The coefficient of the \(y\) term is \(-\text{10}\). Half of the coefficient of the \(y\) term is \(-\text{5}\) and the square of it is \(\text{25}\). Therefore \(y^2 - 10y + 25 - 25 - 11 = 0\).
Write the trinomial as a perfect square
\begin{align*} (y^2 - 10y + 25) - 25 - 11 &= 0 \\ (y-5)^2 - 36 &= 0 \end{align*}Method \(\text{1}\): Take square roots on both sides of the equation
\begin{align*} (y-5)^2 &= 36 \\ y-5 &= \pm\sqrt{36} \end{align*}Important: When taking a square root always remember that there is a positive and negative answer, since \((6)^2 = 36\) and \((-6)^2 = 36\).
\[y - 5 = \pm 6\]Solve for \(y\)
\begin{align*} \text{If } y - 5 &= 6 \\ y &= 11 \\ \text{Or if } y - 5 &= -6 \\ y &= - 1 \end{align*}Therefore \(y = 11\) or \(y = -1\).
Method \(\text{2}\): Factorise the expression as a difference of two squares
\begin{align*} (y-5)^2 - (6)^2 &= 0 \\ \left[\left(y-5\right) + 6\right] \left[\left(y-5\right) - 6\right] &= 0 \end{align*}Simplify and solve for \(y\)
\begin{align*} (y+1)(y-11) &= 0 \\ \therefore y = -1 \text{ or } y &= 11 \end{align*}Write the final answer
\[y = -1 \text{ or } y = 11\]Notice that both methods produce the same answer. These roots are rational because \(\text{36}\) is a perfect square.
Quadratic polynomials
Textbook Exercise 5.2
Solve the following quadratic equations by factorisation. Answers may be left in surd form, where applicable.
\(7p^2+14p=0\)
\begin{align*}
7p^2 + 14p &= 0 \\
7p(p+2) &= 0 \\
p(p+2) &= 0 \\
p = 0 &\text{ or } p = -2
\end{align*}
\({k}^{2}+5k-36=0\)
\begin{align*}
k^2 + 5k - 36 &= 0\\
(k-4)(k+9) &= 0\\
k = 4 &\text{ or } k = -9
\end{align*}
\(400 = 16h^2\)
\begin{align*}
16h^2 - 400 &= 0\\
16(h^2 - 25)&= 0\\
(h-5)(h+5) &= 0\\
h &= \pm 5
\end{align*}
\((x-1)(x+10) + 24 = 0\)
\begin{align*}
(x-1)(x+10) + 24 &= 0 \\
x^2 + 9x - 10 + 24 &= 0 \\
x^2 + 9x + 14 &= 0 \\
(x+7)(x+2) &= 0\\
x = -7 &\text{ or } x=-2
\end{align*}
\(y^2 - 5ky + 4k^2 = 0\)
\begin{align*}
y^2 - 5ky + 4k^2 &= 0\\
(y-4k)(y-k) &= 0\\
y = 4k &\text{ or } y=k
\end{align*}
Solve the following equations by completing the square:
\({p}^{2}+10p-2=0\)
\begin{align*}
p^2 + 10p - 2 &= 0\\
p^2 +10p&= 2\\
p^2 +10p +25 &= 2 + 25\\
(p+5)^2 - 27 &= 0\\
(p+5)^2 &= 27\\
\therefore (p + 5) &= \pm \sqrt{27} \\
\therefore (p+5) = -\sqrt{27} &\text{ or } (p+5) = \sqrt{27}\\
\therefore p = -5 - 3\sqrt{3} &\text{ or } p= -5 + 3\sqrt{3}
\end{align*}
\(2(6y + y^2) = -4\)
\begin{align*}
y^2 +6y &= -2\\
y^2 + 6y + 9&= -2 + 9\\
(y+3)^2 &= 7\\
y + 3 &= \pm \sqrt{7}\\
y &= -3 \pm \sqrt{7}
\end{align*}
\({x}^{2}+5x+9=0\)
\begin{align*}
x^2+5x+9 &= 0\\
x^2 + 5x &= -9\\
x^2 + 5x + \frac{25}{4} &= -9 + \frac{25}{4}\\
\left(x+\frac{5}{2}\right)^2 &= -\frac{11}{4}\\
x + \frac{5}{2} &= \pm \sqrt{-\frac{11}{4}}\\
\text{ No real solution }
\end{align*}
\(f^2 + 30 = 2(10-8f)\)
\begin{align*}
f^2 + 30 &= 2(10 - 8f) \\
f^2+16f+10 &=0\\
f^2 +16f &= -10\\
f^2 + 16f+64 &= -10 +64\\
(f+8)^2 &=54\\
f + 8 &= \pm \sqrt{54}\\
f &= -8 \pm \sqrt{9 \times 6} \\
\therefore f &= -8 \pm 3\sqrt{6}
\end{align*}
\(3{x}^{2}+6x-2=0\)
\begin{align*}
3x^2+6x-2&=0\\
\text{Divide through by } & 3 \text{ to get leading coefficient of } 1 \\
x^2 + 2x &= \frac{2}{3}\\
x^2 + 2x + 1 &= \frac{2}{3} + 1\\
(x+1)^2 &= \frac{5}{3}\\
x +1 &= \pm \sqrt{\frac{5}{3}}\\
x &= -1 \pm \sqrt{\frac{5}{3}}
\end{align*}
Solve the following using the quadratic formula.
\(3{m}^{2}+m-4=0\)
\[a=3 ; \quad b = 1; \quad c = -4\]
\begin{align*}
3m^2 + m - 4 &= 0\\
m &= \dfrac{-(1) \pm \sqrt{1^2 - 4(3)(-4)}}{2(3)}\\
& = \dfrac{-1 \pm \sqrt{1+48}}{6}\\
&=\dfrac{-1 \pm \sqrt{49}}{6}\\
&=\dfrac{-1 \pm 7}{6}\\
m = \dfrac{-1+7}{6}= \frac{6}{6}=1 &\text{ or } m = \dfrac{-1-7}{6}=\frac{-8}{6} = - \frac{4}{3}
\end{align*}
Note: it would be easier and faster to solve by factorisation rather than using the quadratic formula. If the question does not specify which method to use, first try to solve by factorisation.
\(2{t}^{2}+6t+5=0\)
\begin{align*} 2t^2 + 6t + 5 &= 0\\ t&=\dfrac{-6 \pm \sqrt{(6)^2 - 4(2)(5)}}{2(2)}\\ &= \dfrac{-6 \pm \sqrt{36 - 40}}{4}\\ &=\dfrac{-6 \pm \sqrt{-4}}{4}\\ \text{No real solution} \end{align*}
\({y}^{2}-4y+2=0\)
\begin{align*}
y^2 - 4y + 2 &= 0\\
y &= \dfrac{-(-4) \pm \sqrt{(-4)^2 - 4(1)(2)}}{2(1)}\\
&= \dfrac{4 \pm \sqrt{16-8}}{2}\\
&=\dfrac{4 \pm \sqrt{8}}{2}\\
&= 2 \pm \sqrt{2} \\
\therefore y &= 2 + \sqrt{2}\text{ or } y = 2 - \sqrt{2}
\end{align*}
\(3f -2 = -2f^2\)
\begin{align*}
2f^2 + 3f - 2 &= 0\\
f &= \dfrac{-3 \pm \sqrt{3^2-4(2)(-2)}}{2(2)}\\
&= \dfrac{-3 \pm \sqrt{9+16}}{4}\\
&= \dfrac{-3 \pm \sqrt{25}}{4}\\
&= \dfrac{-3 \pm 5}{4}\\
\text{therefore } f =\dfrac{-3 + 5}{4} = \frac{1}{2}&\text{ or } f =\dfrac{-3 - 5}{4} = -2
\end{align*}
Factorise the following:
\(27p^{3} - 1\)
\[27p^{3} - 1 = (3p-1)(9p^{2} + 3p + 1)\]
\(16 + \frac{2}{x^{3}}\)
\begin{align*}
16 + \frac{2}{x^{3}} &= 2 \left( 8 + \frac{1}{x^{3}} \right) \\
&= 2 \left( 2 + \frac{1}{x} \right) \left( 4 - \frac{2}{x} + \frac{1}{x^{2}} \right)
\end{align*}
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5.2 Cubic polynomials
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