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Polynomials, SMB-017

Sketch  `y=-(x+2)(x-1)^2` on the graph below, clearly showing all intercepts.  (3 marks)
 

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Show Worked Solution

`xtext{-axis intercepts at}\ \ x=-2 and 1\ text{(touches only)}`

`ytext{-intercept when}\ \ x=0\ \ =>\ \ y=-2xx(-1)^2=-2`
 

Polynomials, SMB-016

Sketch  `y=(x+2)^2(2x-1)` on the graph below, clearly showing all intercepts .  (3 marks)
 

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Show Worked Solution

`xtext{-axis intercepts at}\ \ x=-2\ text{(touches only) and}\ 1/2`

`ytext{-intercept when}\ x=0\ \ =>\ \ y=2^2(-1)=-4`
 

Polynomials, SMB-015

Sketch  `y=(x+1)^2(x-3)` on the graph below, clearly showing all intercepts .  (3 marks)
 

 

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Show Worked Solution

`xtext{-axis intercepts at}\ \ x=-1\ text{(touches only) and}\ 3`

`ytext{-intercept when}\ x=0\ \ =>\ \ y=1^2(-3)=-3`
 

Polynomials, SMB-014

Sketch  `y=x(x-2)(x+3)` on the graph below, clearly showing all intercepts .  (3 marks)

 

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Show Worked Solution

`xtext{-axis intercepts at}\ \ x=-3, 0 and 2`

`ytext{-intercept at}\ (0,0)`

`text{At}\ \ x=1\ \ =>\ \ y=-4`
  

Polynomials, SMB-013

Sketch  `y=(x+1)(x+2)(x-1)` on the graph below, clearly showing all intercepts .  (3 marks)

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Show Worked Solution

`xtext{-axis intercepts at}\ \ x=-2, -1 and 1`

`ytext{-intercept when}\ \ x=0\ \ =>\ \  y=1xx2xx-1=-2`
  

Polynomials, SMB-012

`h(x)=x^3+3x^2+x-5`.

  1. Show  `h(1)=0`  (1 mark)

  2. Express `h(x)` in the form `h(x)=(x-1)*g(x)` where `g(x)` is a quadratic factor.  (2 marks)

  3. Justify that `h(x)` only has one zero.  (2 marks)

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i.    `text{Proof (See worked solutions)}`

ii.    `h(x)=(x-1)(x^2+4x+5)`

iii.   `text{Proof (See worked solutions)}`

Show Worked Solution

i.   `h(x)=x^3+3x^2+x-5`.

`h(1) = 1+3+1-5=0`
 

ii.   `h(x)=(x-1)*g(x)`

`text{By long division:}`
 

`h(x)=(x-1)(x^2+4x+5)`
 

iii.   `text{Consider the roots of}\ \ y=x^2+4x+5`

`Δ = b^2-4ac=4^2-4*1*5=-4<0`

`text{Since}\ \ Δ<0\ \ =>\ \ text{No zeros (roots)}`

`:. h(x)\ text{only has 1 zero at}\ x=1\ (h(1)=0)`

Polynomials, SMB-011

`g(x)=(x-1)(x^2-2x+8)`.

Justify that `g(x)` only has one zero.  (2 marks)

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`text{Proof (See worked solutions)}`

Show Worked Solution

`g(x)=(x-1)(x^2-2x+8)`.

`g(1)=0\ \ =>\ \ text{one zero at}\ \ x=1`

`text{Consider the roots of}\ \ y=x^2-2x+8`

`Δ = b^2-4ac=(-2)^2-4*1*8=-28<0`

`text{Since}\ \ Δ<0\ \ =>\ \ text{No zeros (roots)}`

`:. g(x)\ text{only has 1 zero}`

Polynomials, SMB-010 MC

`P(x)` is a monic polynomial of degree 4.

The maximum number of zeros that `P(x)` can have is

  1. `0`
  2. `1`
  3. `3`
  4. `4`
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`D`

Show Worked Solution

`text{A polynomial of degree 4 has a leading term}\ ax^4`

`text{A monic polynomial of degree 4 has a leading term}\ x^4`

`:.\ text{Maximum number of zeroes}\ = 4`

`=>D`

Polynomials, SMB-009

Let  `P(x) = x^3+5x^2+2x-8`.

  1. Show that  `P(-2) = 0`.  (1 mark)

  2. Hence, factor the polynomial  `P(x)`  as  `A(x)B(x)`, where  `B(x)`  is a quadratic polynomial.  (2 marks)

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  1. `text(See Worked Solutions)`
  2. `P(x)=(x+2)(x^2+3x-4)`
Show Worked Solution
i.    `P(-2)` `= (-2)^3+ 5(-2)^2+2(-2)-8`
    `=-8+20-4-8`
    `= 0`

 

ii.  `text{Since}\ \ P(-2)=0\ \ =>\ \ (x+2)\ text{is a factor of}\ P(x)`

`P(x)=A(x)B(x)=(x+2)*B(x)`

`text{Using long division:}\ P(x)-:(x+2)=B(x)`
 

`:.P(x)=(x+2)(x^2+3x-4)`

Polynomials, SMB-008

Consider the polynomial  `P(x) = 2x^4+3x^3-12x^2-7x+6`.

Fully factorised, `P(x) = (2x-1)(x+3)(x+a)(x-b)`

Find the value of `a` and `b` where `a,b>0`.  (3 marks)

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`a=1, b=2`

Show Worked Solution

`text{Test for factors (by trial and error):}`

`P(1) = 2+3-12-7+6 = -8`

`P(-1) = 2-3-12+7+6 = 0\ \ =>\ \ (x+1)\ \ text{is a factor}`

`P(2) = 32+24-48-14+6 = 0\ \ =>\ \ (x-2)\ \ text{is a factor}`

`:. a=1, b=2`

Polynomials, SMB-007

Consider the polynomial  `P(x) = 3x^3+x^2-10x-8`.

Is `(x+2)` a factor of `P(x)`? Justify your answer.  (2 marks)

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`P(-2) = -24+4+20-8=-8`

`:. (x+2)\ \ text(is not a factor of)\ P(x)`

Show Worked Solution

`P(-2) = -24+4+20-8=-8`

`:. (x+2)\ \ text(is not a factor of)\ P(x)`

Polynomials, SMB-006

Consider the polynomial  `P(x) = 2x^3-7x^2-7x+12`.

  1. Show that  `(x-1)`  is a factor of  `P(x)`.  (1 mark)

  2. Fully factorise `P(x)`.  (2 marks)

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  1. `text(See Worked Solution)`
  2. `P(x)=(x-1)(2x+3)(x-4)`
Show Worked Solution

i.   `P(1) = 2-7-7+12=0`

`:. (x-1)\ \ text(is a factor of)\ P(x)`

 

ii.   `text{Using part (i)} \ => (x-1)\ text{is a factor of}\ P(x)`

`P(x) = (x-1)*Q(x)`
 

`text(By long division:)`
 

`P(x)` `= (x-1) (2x^2-5x-12)`
  `= (x-1)(2x+3)(x-4)`

Polynomials, SMB-005

Consider the polynomial  `P(x) = x^3-4x^2+x+6`.

  1. Show that  `x = -1`  is a zero of  `P(x)`.  (1 mark)

  2. Find the other zeros.  (2 marks)

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  1. `text(See Worked Solution)`
  2. `x = 2 and 3`
Show Worked Solution

i.   `P(-1) = -1-4-1+6 = 0`

`:. x=-1\ \ text(is a zero)`

 

ii.   `text{Using part (i)} \ => (x+1)\ text{is a factor of}\ P(x)`

`P(x) = (x+1)*Q(x)`
 

`text(By long division:)`

`P(x)` `= (x+1) (x^2-5x+6)`
  `= (x+1)(x-2)(x-3)`

 
`:.\ text(Other zeroes are:)`

`x = 2 and x = 3`

Polynomials, SMB-004

Let  `p(x)=x^{3}-2 a x^{2}+x-1`. When `p(x)` is divided by `(x+2)`, the remainder is 5.

Find the value of `a`.  (2 marks)

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`-2`

Show Worked Solution

`text{Since}\ \ p(x) -: (x+2)\ \ text{has a remainder of 5:}`

`P(-2)` `=5`  
`5` `=(-2)^3-2a(-2)^2-2-1`  
`5` `=-8-8a-2-1`  
`8a` `=-16`  
`:.a` `=-2`  

Functions, EXT1 F2 2022 HSC 3 MC

Let `P(x)` be a polynomial of degree 5. When `P(x)` is divided by the polynomial `Q(x)`, the remainder is `2x+5`.

Which of the following is true about the degree of `Q`?

  1. The degree must be 1.
  2. The degree could be 1.
  3. The degree must be 2.
  4. The degree could be 2.
Show Answers Only

`D`

Show Worked Solution

`text{Given}\ \ P(x)\ \ text{has degree 5}`

`P(x) -: Q(x)\ \ text{has remainder}\ \ 2x+5`

`text{Consider examples to resolve possibilities:}`

`text{eg.}\ \ x^5+2x+5 -: x^3 = x^2+\ text{remainder}\ 2x+5`

`:.\ text{Degree must be 2 is incorrect}`

`Q(x)\ \ text{can have a degree of 2, 3 or 4}`

`=>D`


♦ Mean mark 51%.

Functions, EXT1 F2 2020 HSC 11a

Let  `P(x) = x^3 + 3x^2-13x + 6`.

  1. Show that  `P(2) = 0`.  (1 mark)

  2. Hence, factor the polynomial  `P(x)`  as  `A(x)B(x)`, where  `B(x)`  is a quadratic polynomial.  (2 marks)

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  1. `text(See Worked Solutions)`
  2. `P(x) = (x-2)(x^2 + 5x – 3)`
Show Worked Solution
i.    `P(2)` `= 8 + 12-26 + 6`
    `= 0`

 

ii.   

`:. P(x) = (x-2)(x^2 + 5x – 3)`

Functions, EXT1 F2 2019 HSC 11d

Find the polynomial  `Q(x)`  that satisfies  `x^3 + 2x^2-3x-7 = (x-2) Q(x) + 3`.  (2 marks)

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`Q(x ) = x^2 + 4x + 5`

Show Worked Solution
`(x-2) ⋅ Q(x) + 3` `= x^3 + 2x^2-3x-7`
`(x-2) ⋅ Q(x)` `= x^3 + 2x^2-3x-10`

 

`:. Q(x ) = x^2 + 4x + 5`

Functions, EXT1 F2 2018 HSC 11a

Consider the polynomial  `P(x) = x^3-2x^2-5x + 6`.

  1. Show that  `x = 1`  is a zero of  `P(x)`.  (1 mark)

  2. Find the other zeros.  (2 marks)

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  1. `text(See Worked Solution)`
  2. `x = -2 and x = 3`
Show Worked Solution

i.   `P(1) = 1-2-5 + 6 = 0`

`:. x=1\ \ text(is a zero)`

 

ii.   `text{Using part (i)} \ => (x-1)\ text{is a factor of}\ P(x)`

`P(x) = (x-1)*Q(x)`
 

`text(By long division:)`

`P(x)` `= (x-1) (x^2-x-6)`
  `= (x-1) (x-3) (x + 2)`

 
`:.\ text(Other zeroes are:)`

`x = -2 and x = 3`

Functions, EXT1 F2 2007 HSC 2c

The polynomial  `P(x) = x^2 + ax + b`  has a zero at  `x = 2`. When  `P(x)`  is divided by  `x + 1`, the remainder is `18`.

Find the values of  `a`  and  `b`.  (3 marks)

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`a = -7\ \ text(and)\ \ b = 10`

Show Worked Solution

`P(x) = x^2 + ax + b`

`text(S)text(ince there is a zero at)\ \ x = 2,`

`P(2)` `=0`  
`2^2 + 2a + b` `= 0`  
`2a + b` `= -4`       `…\ (1)`

 
`P(-1) = 18,`

`(-1)^2-a + b` `= 18`  
`-a + b` `= 17`    `…\ (2)`

 
`text(Subtract)\ \ (1)-(2),`

`3a` `= -21`
`a` `= -7`

 
`text(Substitute)\ \ a = -7\ \ text{into (1),}`

`2(-7) + b` `= -4`
`b` `= 10`

 
`:.a = -7\ \ text(and)\ \ b = 10`

Functions, EXT1 F2 2015 HSC 11f

Consider the polynomials  `P(x) = x^3-kx^2 + 5x + 12`  and  `A(x) = x - 3`.

  1. Given that  `P(x)`  is divisible by  `A(x)`, show that  `k = 6`.  (1 mark)

  2. Find all the zeros of  `P(x)`  when  `k = 6`.  (2 marks)

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  1. `text(See Worked Solutions)`
  2. `3, 4, −1`
Show Worked Solution
i.    `P(x)` `= x^3-kx^2 + 5x + 12`
  `A(x)` `= x-3`

 
`text(If)\ P(x)\ text(is divisible by)\ A(x)\ \ =>\ \ P(3) = 0`

`3^3-k(3^2) + 5 xx 3 + 12` `= 0`
`27-9k + 15 + 12` `= 0`
`9k` `= 54`
`:.k` `= 6\ \ …\ text(as required)`

 

ii.  `text(Find all roots of)\ P(x)`

`P(x)=(x-3)*Q(x)`

`text{Using long division to find}\ Q(x):`
 

`:.P(x)` `= x^3-6x^2 + 5x + 12`
  `= (x-3)(x^2-3x − 4)`
  `= (x-3)(x-4)(x + 1)`

 
`:.\ text(Zeros at)\ \ \ x = -1, 3, 4`

Functions, EXT1 F2 2008 HSC 1a

The polynomial  `x^3`  is divided by  `x + 3`. Calculate the remainder.   (2 marks)

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`-27`

Show Worked Solution
`P(-3)` `= (-3)^3`
  `= -27`

 
`:.\ text(Remainder when)\ x^3 -: (x + 3) = -27`

MARKER’S COMMENT: “Grave concern” that many who found `P(-3)=-27` stated the remainder was 27.

Functions, EXT1 F2 2014 HSC 9 MC

The remainder when the polynomial  `P(x) = x^4-8x^3-7x^2 + 3`  is divided by  `x^2 + x`  is  `ax + 3`.

What is the value of  `a`?

  1. `-14`
  2. `-11`
  3. `-2`
  4. `5`
Show Answers Only

`C`

Show Worked Solution

`P(x) = x^4-8x^3-7x^2 + 3`

`text(Given)\ \ P(x)` `= (x^2 + x) *Q(x) + ax + 3`
  `= x (x + 1) Q(x) + ax + 3`

 
`P(-1) = 1 + 8-7 + 3 = 5`

`:. -a + 3` `= 5`
`a` `= -2`

 
`=>  C`

Functions, EXT1 F2 2009 HSC 2a

The polynomial  `p(x) = x^3-ax + b`  has a remainder of  `2`  when divided by  `(x-1)`  and a remainder of  `5`  when divided by  `(x + 2)`.  

Find the values of  `a`  and  `b`.   (3 marks)

Show Answers Only
`a` `= 4`
`b` `= 5`
Show Worked Solution
`p(x)` `= x^3-ax + b`
`P(1)` `= 2`
`1-a + b` `= 2`
`b` `= a+1\ \ \ …\ text{(1)}`
`P (-2)` `= 5`
`-8 + 2a + b` `= 5`
`2a + b` `= 13\ \ \ …\ text{(2)}`

 

`text(Substitute)\ \ b = a+1\ \ text(into)\ \ text{(2)}`

`2a + a+1` `= 13`
`3a` `= 12`
`:. a` `= 4`
`:. b` `= 5`

Functions, EXT1 F2 2013 HSC 1 MC

The polynomial  `P(x) = x^3-4x^2-6x + k`  has a factor  `x-2`.

What is the value of  `k`? 

  1. `2` 
  2. `12`
  3. `20` 
  4. `36`  
Show Answers Only

`C`

Show Worked Solution

`P(x) = x^3-4x^2-6x  + k`

`text(S)text(ince)\ \ (x-2)\ \ text(is a factor,)\ \ P(2) = 0`

`2^3-4*2^2-6*2 + k` `= 0`
`8-16-12 + k` `= 0`
`k` `= 20`

 
`=>  C`

Functions, EXT1 F2 2010 HSC 2c

Let  `P(x) = (x + 1)(x-3) Q(x) + ax + b`, 

where  `Q(x)`  is a polynomial and  `a`  and  `b`  are real numbers.

The polynomial  `P(x)`  has a factor of  `x-3`.

When  `P(x)`  is divided by  `x + 1`  the remainder is  `8`. 

  1. Find the values of  `a`  and  `b`.  (2 marks)

  2. Find the remainder when  `P(x)`  is divided by  `(x + 1)(x-3)`.     (1 mark)

Show Answers Only
  1. `a = -2,\ b = 6`
  2. ` -2x + 6`
Show Worked Solution

i.  `P(x) = (x+1)(x-3)Q(x) + ax + b`

`(x-3)\ \ text{is a factor   (given)}`

`:. P (3)` `= 0`
`3a + b` `= 0\ \ \ …\ text{(1)}`

 
`P(x) ÷ (x+1)=8\ \ \ text{(given)}`

`:.P(-1)` `= 8`
`-a + b` `= 8\ \ \ …\ text{(2)}`

 
`text{Subtract  (1) – (2)}`

`4a` `= -8`
`a` `= -2`

 
`text(Substitute)\ \ a = -2\ \ text{into (1)}`

`-6 + b` `= 0`
`b` `= 6`

 
`:. a= – 2, \ b=6` 
 

ii.  `P(x) -: (x + 1)(x-3)`

`= ((x+1)(x-3)Q(x)-2x + 6)/((x+1)(x-3))`

`= Q(x) + (-2x + 6)/((x+1)(x-3))`

 
`:.\ text(Remainder is)\ \ -2x + 6`

COMMENT: This question requires a fundamental understanding of the remainder theorem.

Functions, EXT1 F2 2011 HSC 2a

Let  `P(x) = x^3-ax^2 + x`  be a polynomial, where  `a`  is a real number.

When  `P(x)`  is divided by  `x-3`  the remainder is  `12`.

Find the remainder when  `P(x)`  is divided by  `x + 1`.    (3 marks)

Show Answers Only

`-4`

Show Worked Solution

`P(x) = x^3 – ax^2 + x`

`text(S)text(ince)\ \ P(x) -: (x – 3)\ \ text(has remainder 12,)`

`P(3) = 3^3-a xx 3^2 + 3` `=12`
`27-9a + 3` `= 12`
`9a` `= 18`
`a` `=2`

 
`:.\ P(x) = x^3-2x^2 + x`

 

`text(Remainder)\ \ P(x) -: (x + 1)\ \ text(is)\ \ P(–1)`

`P(-1)` `= (-1)^3-2(-1)^2-1`
  `= – 4`

Functions, EXT1 F2 2012 HSC 8 MC

When the polynomial  `P(x)`  is divided by  `(x + 1)(x-3)`, the remainder is  `2x + 7`.  

What is the remainder when  `P(x)`  is divided by  `x-3`? 

  1. `1` 
  2. `7` 
  3. `9` 
  4. `13` 
Show Answers Only

`D`

Show Worked Solution

`text(Let)\ \ P(x) =A(x) * Q(x) + R(x)`

`text(where)\ \ A(x) = (x + 1)(x-3),\ text(and)\ \ R(x)=2x+7`

`text(When)\ \ P(x) -: (x-3),\ text(remainder) = P(3)`

`P(3)` `= 0 + R(3)`
  `= (2 xx 3) + 7`
  `= 13`

 
`=>  D`