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Functions, EXT1 F2 2023 HSC 11c

Consider the polynomial

\(P(x)=x^3+a x^2+b x-12\),

where \(a\) and \(b\) are real numbers.

It is given that  \(x+1\)  is a factor of \(P(x)\) and that, when \(P(x)\) is divided by  \(x-2\), the remainder is \(-18\) .

Find \(a\) and \(b\).  (3 marks)

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\(a=2, \ b=-11\)

Show Worked Solution

\((x+1)\ \ \text{is a factor of}\ P(x)\ \ \Rightarrow\ \ P(-1)=0 \)

\(0\) \(=(-1)^3+a-b-12\)  
\(a-b\) \(=13\ \ …\ (1) \)  

 
\(P(x) ÷ (x-2) \ \text{has a remainder of}\ -18 \ \Rightarrow\ \ P(2)=-18 \)

\(-18\) \(=8+4a+2b-12\)  
\(4a+2b\) \(=-14\)  
\(2a+b\) \(=-7\ \ …\ (2) \)  

 
\(\text{Add}\ \ (1) + (2): \)

\(3a=6\ \ \Rightarrow\ \ a=2\)

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

\(2-b=13\ \ \Rightarrow\ \ b=-11 \)

\(\therefore a=2, \ b=-11\)

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 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 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`  
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`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)

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  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.