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Polynomials, EXT1 EQ-Bank 11

The polynomial  \(R(x)=x^3+p x^2+q x+6\)  has a double zero at  \(x=-1\)  and a zero at  \(x=s\).

Find the values of \(p, q\) and \(s\).   (3 marks)

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\(s=-6, \ p=8, \ q=13\)

Show Worked Solution

\(R(x)=x^3+p x^2+q x+6\)

\(R(x)\ \text{is monic with a zero at} \ s \ \text{and double zero at}\ -1:\)

\(R(x)\) \(=(x+1)^2(x-s)\)
  \(=\left(x^2+2 x+1\right)(x-s)\)
  \(=x^3+2 x^2+x-s x^2-2 s x-s\)
  \(=x^3+(2-s) x^2+(1-2 s) x-s\)

 

\(\text{Equating coefficients:}\)

\(-s=6 \ \Rightarrow \ s=-6\)

\(p=2-(-6)=8\)

\(q=1-2(-6)=13\)

Polynomials, EXT1 EQ-Bank 10

The polynomial  \(R(x)=2 x^4+a x^3+b x^2+c x+d\)  has a double zero at  \(x=1\), a zero at  \(x=-3\), and passes through the point \((0,-12)\).

Find the integer values of \(a, b, c, d\) and the fourth zero of the polynomial.   (4 marks)

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\(a=-2, \ b=-14, \ c=26, \ d=-12\)

\(\text{Fourth zero:} \ \ x=2\)

Show Worked Solution

\(R(x)=2 x^4+a x^3+b x^2+c x+d\)

\(\text{Since leading coefficient is 2 with a double zero at 1 and a zero at }-3:\)

\(R(x)=2(x-1)^2(x+3)(x-k) \ \ \text{where} \ k \ \text{is the fourth zero.}\)

\(\text{The polynomial passes through}\ (0,-12):\)

\(R(0)=2(0-1)^2(0+3)(0-k)=-12\ \ \Rightarrow\ \ k=2\)
 

\(\text{Expanding}\ R(x):\)

\(R(x)\) \(=2(x-1)^2(x+3)(x-2)\)  
  \(=2\left(x^2-2 x+1\right)(x+3)(x-2) \)  
  \(=2(x^3+3 x^2-2 x^2-6 x+x+3)(x-2) \)  
  \(=2(x^3+x^2-5 x+3)(x-2) \)  
  \(=2(x^4+x^3-5 x^2+3 x-2 x^3-2 x^2+10 x-6) \)  
  \(=2 x^4-2 x^3-14 x^2+26 x-12\)  

 

\(\text{Equating coefficients:}\)

\(a=-2, \ b=-14, \ c=26, \ d=-12\)

\(\text{Fourth zero:} \ \ x=2\)

Polynomials, EXT1 EQ-Bank 9

A polynomial has the equation

\(Q(x)=(x+1)^2(x-2)\left(x^2+2 x-8\right)\)

  1. Express  \(Q(x)\)  as a product of linear factors and determine the multiplicity of each of its roots.   (2 marks)

  2. Hence, without using calculus, draw a sketch of \(y=Q(x)\), showing all  \(x\)-intercepts.   (2 marks)

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a.    \(Q(x) = (x+1)^2(x-2)\left(x^2+2 x-8\right) \)

\(\text{Roots:}\)

\(x=-1 \ \ \text{(multiplicity 2)}\)

\(x=2 \ \ \text{(multiplicity 2)}\)

\(x=4 \ \ \text{(multiplicity 1)}\)
 

b.

Show Worked Solution
a.     \(Q(x)\) \(=(x+1)^2(x-2)\left(x^2+2 x-8\right)\)
    \(=(x+1)^2(x-2)(x-2)(x+4)\)
    \(=(x+1)^2(x-2)^2(x-4)\)

 

\(\text{Roots:}\)

\(x=-1 \ \ \text{(multiplicity 2)}\)

\(x=2 \ \ \text{(multiplicity 2)}\)

\(x=4 \ \ \text{(multiplicity 1)}\)
 

b.    \(Q(x) \ \text{degree}=5, \ \text{Leading coefficient}=1\)

\(\text{As} \ \ x \rightarrow-\infty, y \rightarrow-\infty\)

\(\text{As} \ \ x \rightarrow \infty, y \rightarrow \infty\)

\(\text{At}\ \ x=0, \ y=1^2 \times (-2)^2 \times -4 = -16\)

Polynomials, EXT1 EQ-Bank 8

A polynomial \(f(x)\) is defined by

\(f(x)=-3x^3+27x^2+12x-1\)

Explain what happens to \(f(x)\) as  \(x \rightarrow-\infty\).   (2 marks)

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\(\text{Since}\ f(x) \ \text{has degree 3:}\)

\(\text{As} \ \ x \rightarrow-\infty, x^3 \rightarrow-\infty\)

\(f(x) \ \text{has leading coefficient}=-3\)

\(\therefore\ \text{As} \ \ x \rightarrow-\infty,-3 x^3 \rightarrow \infty, \ f(x) \rightarrow \infty\)

Show Worked Solution

\(\text{Since}\ f(x) \ \text{has degree 3:}\)

\(\text{As} \ \ x \rightarrow-\infty, x^3 \rightarrow-\infty\)

\(f(x) \ \text{has leading coefficient}=-3\)

\(\therefore\ \text{As} \ \ x \rightarrow-\infty,-3 x^3 \rightarrow \infty, \ f(x) \rightarrow \infty\)

Polynomials, EXT1 EQ-Bank 7

Consider a polynomial  \(y=(x+3)^2(2-x) \cdot Q(x)\)

where  \(Q(x)=6-x-x^2\)

Explain what happens to \(y\) as  \(x \rightarrow-\infty\).   (2 marks)

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\(y\) \(=(x+3)^2(2-x)\left(6-x-x^2\right)\)
  \(=(x+3)^2(2-x)(x+3)(2-x)\)
  \(=(x+3)^3(2-x)^2\)

 

\(\text{Degree\(=5\), Leading co-efficient\(=1\)}\)

\(\therefore \text{As} \ \ x \rightarrow-\infty, x^5 \rightarrow-\infty, y \rightarrow-\infty\)

Show Worked Solution
\(y\) \(=(x+3)^2(2-x)\left(6-x-x^2\right)\)
  \(=(x+3)^2(2-x)(x+3)(2-x)\)
  \(=(x+3)^3(2-x)^2\)

 

\(\text{Degree\(=5\), Leading co-efficient\(=1\)}\)

\(\therefore \text{As} \ \ x \rightarrow-\infty, x^5 \rightarrow-\infty, y \rightarrow-\infty\)

Polynomials, EXT1 EQ-Bank 5

Consider the function  \(P(x)=(x-1)^2(x+2)\left(x^2+3 x-4\right)\)

  1. By expressing \(P(x)\) as a product of its linear factors, identify its zeroes and the multiplicity of each zero.   (2 marks)

  2. Without using calculus, draw a sketch of  \(y=P(x)\)  showing any \(x\)-intercepts.   (2 marks)

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a.    \(P(x)=(x-1)^3(x+2)(x+4)\)

\(\text{Roots:}\)

\(x=-2 \ \ \text{(multiplicity 1)}\)

\(x=-4 \ \ \text{(multiplicity 1)}\)

\(x=1 \ \ \text{(multiplicity 3)}\)
 

b.  

Show Worked Solution
a.     \(P(x)\) \(=(x-1)^2(x+2)\left(x^2+3 x-4\right)\)
    \(=(x-1)^2(x+2)(x-1)(x+4)\)
    \(=(x-1)^3(x+2)(x+4)\)

 
\(\text{Roots:}\)

\(x=-2 \ \ \text{(multiplicity 1)}\)

\(x=-4 \ \ \text{(multiplicity 1)}\)

\(x=1 \ \ \text{(multiplicity 3)}\)
 

b.    \(\text{Degree} \ P(x)=5, \ \ \text {Leading coefficient }=1\)

\(\text{As} \ \ x \rightarrow \infty, \ y \rightarrow \infty\)

\(\text{As} \ \ x \rightarrow -\infty, \ y \rightarrow -\infty\)

\(\text{At} \ \ x=0, y=-8\)
 

Polynomials, EXT1 EQ-Bank 5 MC

The graph of  `y = f(x)`  is shown.
 

Which of the following could be the equation of this graph?

  1. `y = (1-x)(2 + x)^3`
  2. `y = (x + 1)(x-2)^3`
  3. `y = (x + 1)(2-x)^3`
  4. `y = (x-1)(2 + x)^3`
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`C`

Show Worked Solution

`text(By elimination:)`

`text(A single negative root occurs when)\ \ x =–1`

`->\ text(Eliminate A and D)`

`text(When)\ \ x = 0, \ y > 0`

`->\ text(Eliminate B)`

`=> C`

Polynomials, EXT1 EQ-Bank 4 MC

Which of the following best represents the graph of  \(y=-5 x(x-2)(3-x)\)?
 

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\(C\)

Show Worked Solution

\(\text{By elimination:}\)

\(\text{Degree = 3,  Leading co-efficient}\ = 5\)

\(\text{As}\ \ x \rightarrow \infty,\ \ y \rightarrow \infty\ \text{(eliminate A and B)}\)

\(\text{When}\ x=1:\)

\(y=-5(-1)(2)=10>0\ \ \text{(eliminate D)}\)

\(\Rightarrow C\)

Polynomials, EXT1 EQ-Bank 3

Consider the polynomial \(P(x)=x(3-x)^3\).

  1. State the degree of the polynomial and identify the leading coefficient.   (1 mark)

  2. Explain what happens to \(y\) as  \(x \rightarrow \pm \infty\).   (1 mark)

  3. Without using calculus, sketch \(P(x)\) showing its general form and any \(x\)-intercepts.   (2 marks)

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a.    \(\text{Degree}\ P(x)=4\)

\(\text{Leading co-efficient}=-1\)
 

b.    \(\text{As} \ \ x \rightarrow-\infty,-x^4 \rightarrow-\infty, \ y \rightarrow-\infty\).

\(\text{As} \ \ x \rightarrow \infty,-x^4 \rightarrow-\infty, \ y \rightarrow-\infty\).
 

c.    \(P(x)\ \text{has zeroes at}\ \ x=0, 3:\)


       

Show Worked Solution

a.    \(\text{Degree}\ P(x)=4\)

\(\text{Leading co-efficient}=-1\)
 

b.    \(\text{As} \ \ x \rightarrow-\infty,-x^4 \rightarrow-\infty, \ y \rightarrow-\infty\).

\(\text{As} \ \ x \rightarrow \infty,-x^4 \rightarrow-\infty, \ y \rightarrow-\infty\).
 

c.    \(P(x)\ \text{has zeroes at}\ \ x=0, 3:\)


       

Polynomials, EXT1 EQ-Bank 4

The polynomial  \(p(x) = x^3 + ax^2 + b\)  has a zero at \(r\) and a double zero at 4.

Find the values of \(a, b\) and \(r\).   (3 marks)

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\(a =-6, b = 32, r = -2\)

Show Worked Solution

\(p(x) = x^3 + ax^2 + b\)

\(\text{Zero at \(r\) and double zero at 4:}\)

\(p(x)\) \(=(x-4)^2(x-r) \)  
  \(=(x^2-8x+16)(x-r)\)  
  \(=x^3-8x^2+16x-rx^2+8rx-16r\)  
  \(=x^3+(-8-r)x^2+(16+8r)x-16r\)  

 

\(\text{Equating coefficients:}\)

\(16+8r=0\ \ \Rightarrow \ \ r=-2\)

\(a=-8-(-2)=-6\)

\(b=-16 \times -2=32\)

Polynomials, EXT1 EQ-Bank 2

Consider a polynomial  \(y=-2 x^5-26 x^4-x+1\).

With reference to the leading term, explain what happens to \(y\) as  \(x \rightarrow-\infty\).   (2 marks)

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\(\text{As}\ \ x \rightarrow-\infty,\ \ x^{5}\ \rightarrow-\infty\)

\(\text{Since the leading coefficient \((-2)\) is negative:}\)

\(\text{As}\ \ x \rightarrow-\infty,\ \ y \rightarrow \infty.\)

Show Worked Solution

\(\text{As}\ \ x \rightarrow-\infty,\ \ x^{5}\ \rightarrow-\infty\)

\(\text{Since the leading coefficient \((-2)\) is negative:}\)

\(\text{As}\ \ x \rightarrow-\infty,\ \ y \rightarrow \infty.\)

Polynomials, EXT1 EQ-Bank 1

A polynomial has the equation

\(P(x)=(x-1)(x-3)(x+2)^2\left(x^2-x-6\right)\).

  1. By expressing \(P(x)\) as a product of its linear factors, determine the multiplicity of each of the roots of  \(P(x)=0\).   (2 marks)

  2. Hence, without using calculus, draw a sketch of  \(y=P(x)\)  showing all \(x\)-intercepts.   (2 marks)

Show Answers Only
a.     \(P(x)\) \(=(x-1)(x-3)(x+2)^2\left(x^2-x-6\right)\)
    \(=(x-1)(x-3)(x+2)^2(x-3)(x+2)\)
    \(=(x-1)(x-3)^2(x+2)^3\)

 

\(\text{Roots:}\)

\(x=1\ \text{(multiplicity 1)}\)

\(x=-2\ \text{(multiplicity 3)}\)

\(x=3\ \text{(multiplicity 2)}\)
 

b.
     

Show Worked Solution
a.     \(P(x)\) \(=(x-1)(x-3)(x+2)^2\left(x^2-x-6\right)\)
    \(=(x-1)(x-3)(x+2)^2(x-3)(x+2)\)
    \(=(x-1)(x-3)^2(x+2)^3\)

 

\(\text{Roots:}\)

\(x=1\ \text{(multiplicity 1)}\)

\(x=-2\ \text{(multiplicity 3)}\)

\(x=3\ \text{(multiplicity 2)}\)
 

b.
     

Functions, EXT1 F2 2020 HSC 5 MC

A monic polynomial `p(x)` of degree 4 has one repeated zero of multiplicity 2 and is divisible by  `x^2 + x + 1`.

Which of the following could be the graph of `p(x)`?

A. B.
C. D.
Show Answers Only

`C`

Show Worked Solution

`text(S)text(ince)\ p(x)\ text(is monic,)`

`=> p(x) = (x-a)^2(x^2 + x + 1)`
 

`text(Consider the factor)\ \ (x^2 + x + 1):`

`Delta = sqrt(1^2-4 · 1 · 1) = sqrt(-3) < 0 =>\ text(No roots)`

`:. text(Only root is)\ \ x = a\ \ (text(multiplicity 2))`

`=>\ text(Eliminate)\ \ B and D`
 

`\text{Since}\ p(x)\ \text{is monic:}`

`text(As)\ \ x -> ∞, \ p(x) -> ∞`

`=>C`