SmarterEd

Aussie Maths & Science Teachers: Save your time with SmarterEd

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)

Show Answers Only

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-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)

Show Answers Only
  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-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)

Show Answers Only
  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)`

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)

Show Answers Only
  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 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)

Show Answers Only
  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`