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Functions, EXT1 F2 2025 HSC 14e

It is given that \(\tan \alpha, \tan \beta\) and \(\tan \gamma\) are the three real roots of the polynomial equation  \(x^3+b x^2+c x-1+b+c=0\), where \(b\) and \(c\) are real numbers and \(c \neq 1\).

Find the smallest positive value of  \(\alpha+\beta+\gamma\).   (3 marks)

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\(\alpha+\beta+\gamma=\dfrac{3 \pi}{4}\)

Show Worked Solution

\(x^3+b x^2+c x-1+b+c=0\)

\(\text{Roots:} \ \tan \alpha, \tan \beta, \tan \gamma\)

\(\tan \alpha+\tan \beta+\tan \gamma=-\dfrac{b}{a}=-b\)

\(\tan \alpha \cdot \tan \beta+\tan \beta \cdot \tan \gamma+\tan \alpha \cdot \tan \gamma=c\)

\(\tan \alpha \cdot \tan \beta \cdot \tan \gamma=-\dfrac{d}{a}=1-b-c\)

♦♦♦ Mean mark 26%.

\(\text { Find smallest +ve value of} \ \ \alpha+\beta+\gamma:\)

\(\tan (\alpha+\beta+\gamma)\) \(=\dfrac{\tan (\alpha+\beta)+\tan \gamma}{1-\tan (\alpha+\beta) \tan \gamma}\)
  \(=\dfrac{\dfrac{\tan \alpha+\tan \beta}{1-\tan \alpha \cdot \tan \beta}+\tan \gamma}{1-\dfrac{\tan \alpha+\tan \beta}{1-\tan \alpha \cdot \tan \beta} \times \tan \gamma}\)
  \(=\dfrac{\tan \alpha+\tan \beta+\tan \gamma(1-\tan \alpha \cdot \tan \beta)}{1-\tan \alpha \cdot \tan \beta-(\tan \alpha+\tan \beta) \tan \gamma}\)
  \(=\dfrac{\tan \alpha+\tan \beta+\tan \gamma-\tan \alpha \cdot \tan \beta \cdot \tan \gamma}{1-(\tan \alpha \cdot \tan \beta+\tan \beta \cdot \tan \gamma+\tan \alpha \cdot \tan \gamma)}\)
  \(=\dfrac{-b-(1-b-c)}{1-c}\)
  \(=\dfrac{-1+c}{1-c}\)
  \(=-1\)

 
\(\therefore \ \text{Smallest +ve value of} \ \ \alpha+\beta+\gamma=\dfrac{3 \pi}{4}\)

Functions, EXT1 F2 2025 HSC 11f

The roots of  \(2 x^3+6 x^2+x-1=0\)  are  \(\alpha, \beta\)  and  \(\gamma\).

What is the value of  \(\dfrac{1}{\alpha \beta}+\dfrac{1}{\alpha \gamma}+\dfrac{1}{\beta \gamma}\) ?   (2 marks)

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\(\dfrac{1}{\alpha \beta} + \dfrac{1}{\alpha \gamma}+\dfrac{1}{\beta \gamma}=-6\)

Show Worked Solution

\(2 x^3+6 x^2+x-1=0\)

\(\dfrac{1}{\alpha \beta}+\dfrac{1}{\alpha \gamma}+\dfrac{1}{\beta \gamma}=\dfrac{\alpha+\beta+\gamma}{\alpha \beta \gamma}\)

\(\alpha+\beta+\gamma=-\dfrac{b}{a}=-3\)

\(\alpha \beta \gamma=-\dfrac{d}{a}=\dfrac{1}{2}\)

\(\therefore \dfrac{1}{\alpha \beta} + \dfrac{1}{\alpha \gamma}+\dfrac{1}{\beta \gamma}=\dfrac{-3}{\frac{1}{2}}=-6\)

Functions, EXT1 F1 F2 EQ-Bank 27

The polynomial  \(P(x)=x^3+5x^2-12x-36\)  has three distinct roots, where one root is the sum of the other two roots.

Find the values of the three roots.   (3 marks)

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\(\text{Roots: \(-6,-2\) and \(3\).}\)

Show Worked Solution

\(P(x)=x^3+5x^2-12x-36\)

\(\text{Let roots}=\alpha, \beta, \alpha\beta\)

\(\alpha+\beta+\alpha\beta=-\dfrac{b}{a}=-5\ \ldots\ (1)\) 

\(\alpha \beta+ \alpha^2 \beta + \alpha \beta^2\) \(=\dfrac{c}{a}=-12\)  
\(\alpha \beta(1+\alpha + \beta)\) \(=-12\ \ldots\ (2)\)   

\(\alpha \times \beta \times \alpha\beta\) \(=-\dfrac{d}{a}=36\)
\(\alpha \beta\) \(= \pm 6\)

 
\(\text{If}\ \ \alpha \beta = 6\ \ \Rightarrow\ \ \alpha+\beta=-11\ \ \text{(using (1) above)}\)

\(\text{Substituting}\ \ \alpha \beta=6\ \ \text{into (2):}\)

\(6(1+\alpha + \beta)= 6(1-11)=-60 \neq -12\ × \)
 

\(\text{If}\ \alpha \beta = -6\ \ \Rightarrow\ \ \alpha+\beta=1\ \text{(using (1) above)}\)

\(\text{Substituting}\ \ \alpha \beta=-6\ \ \text{into (2):}\)

\(-6(1+1)= -12\ \checkmark \)
 

\(\text{We know:}\ \ \alpha + \beta = 1\ \ \text{and}\ \ \alpha \times \beta = -6\)

\(=> \alpha = 3,\ \ \beta =-2\)

\(\therefore\ \text{Roots are:}\ -6, -2, 3\)

Functions, EXT1 F2 2024 HSC 1 MC

The polynomial  \(x^{3} + 2x^{2}-5x-6\)  has zeros  \(-1, -3\) and \(\alpha\).

What is the value of \(\alpha\)?

  1. \(-2\)
  2. \(2\)
  3. \(3\)
  4. \(6\)
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\(B\)

Show Worked Solution

\(\alpha \beta \gamma = -\dfrac{\text{d}}{\text{a}} = 6\)

\(-1 \times -3 \times \alpha\) \(=6\)  
\(\alpha\) \(=2\)  

 
\(\Rightarrow B\)

Functions, EXT1 F2 EQ-Bank 22

In the equation  \(x^3-8 x^2+11x+20=0\)  one of the roots is equal to the sum of the other two roots.

Find the value of the three roots.   (3 marks)

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\(\text{Roots}\ = -1, 4, 5\)

Show Worked Solution

\(x^3-8 x^2+11x+20=0\)

\(\text{Sum of roots}\ =-\dfrac{b}{a}:\)

\(\alpha+\beta+(\alpha + \beta)\) \(=8\)  
\(2\alpha + 2\beta\) \(=8\)  
\(\beta\) \(=4-\alpha\ \ …\ (1)\)  

 
\(\text{Product of roots}\ =-\dfrac{d}{a}:\)

\(\alpha \beta(\alpha + \beta)=-20\ \ …\ (2) \)
 

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

\(\alpha(4-\alpha)(4)\) \(=-20\)  
\(16\alpha-4\alpha^{2}\) \(=-20\)  
\(\alpha^{2}-4\alpha-5\) \(=0\)  
\((\alpha-5)(\alpha+1)\) \(=0\)  

\(\alpha = 5\ \ \text{or}\ -1\)

\(\beta = 5\ \ \text{or}\ -1\ \ \text{(using (1))}\)

\(\alpha+ \beta = 5-1=4\)

\(\therefore \text{Roots}\ = -1, 4, 5\)

Functions, EXT1 F2 2021 HSC 11h

The roots of  `x^4-3x + 6 = 0`  are  `alpha, beta, gamma` and `delta`.

What is the value of  `1/alpha + 1/beta + 1/gamma + 1/delta`?  (2 marks)

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`1/2`

Show Worked Solution

`alpha beta gamma delta = 6/1 = 6`

`alpha beta gamma + beta gamma delta + gamma delta alpha + delta alpha beta = -((-3))/1 = 3`

`1/alpha + 1/beta + 1/gamma + 1/delta= (beta gamma delta + alpha gamma delta + beta gamma delta + alpha beta gamma)/(alpha beta gamma delta)= 1/2`

Trigonometry, EXT1 T3 2020 HSC 14b

  1. Show that  `sin^3 theta-3/4 sin theta + (sin(3theta))/4 = 0`.   (2 marks)

  2. By letting  `x = 4sin theta`  in the cubic equation  `x^3-12x + 8 = 0`.

     

    Show that  `sin (3theta) = 1/2`.   (2 marks)

  3. Prove that  `sin^2\ pi/18 + sin^2\ (5pi)/18 + sin^2\ (25pi)/18 = 3/2`.   (3 marks)

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  1. `text(See Worked Solutions)`
  2. `text(See Worked Solutions)`
  3. `text(See Worked Solutions)`
Show Worked Solution

i.   `text(Prove:)\  \ sin^3 theta-3/4 sin theta + (sin(3theta))/4 = 0`

`text(LHS)` `= sin^3 theta-3/4 sin theta + 1/4 (sin 2thetacostheta + cos2thetasintheta)`
  `= sin^3 theta-3/4 sintheta + 1/4(2sinthetacos^2theta + sintheta(1-2sin^2theta))`
  `= sin^3theta-3/4 sintheta + 1/4(2sintheta(1-sin^2theta) + sintheta – 2sin^3theta)`
  `= sin^3theta-3/4 sintheta + 1/4(2sintheta-2sin^3theta + sintheta-2sin^3theta)`
  `= sin^3theta-3/4sintheta + 3/4sintheta-sin^3theta`
  `= 0`

 

ii.   `text(Show)\ \ sin(3theta) = 1/2`

`text{Using part (i):}`

`(sin(3theta))/4` `= 3/4 sintheta-sin^3 theta`
`sin(3theta)` `= 3sintheta-4sin^3theta\ …\ (1)`

 
`x^3-12x + 8 = 0`

`text(Let)\ \ x = 4 sin theta`

`(4sintheta)^3-12(4sintheta) + 8` `= 0`
`64sin^3theta-48sintheta` `= 0`
`−16underbrace{(3sintheta-4sin^2theta)}_text{see (1) above}` `= −8`
`-16 sin(3theta)` `= −8`
`sin(3theta)` `= 1/2`
♦♦♦ Mean mark (iii) 21%.

 

iii.   `text(Prove:)\ \ sin^2\ pi/18 + sin^2\ (5pi)/18 + sin^2\ (25pi)/18 = 3/2`

`text(Solutions to)\ \ x^3-12x + 8 = 0\ \ text(are)`

`x = 4sintheta\ \ text(where)\ \ sin(3theta) = 1/2`

`text(When)\ \ sin3theta = 1/2,`

`3theta` `= pi/6, (5pi)/6, (13pi)/6, (17pi)/6, (25pi)/6, (29pi)/6, …`
`theta` `= pi/18, (5pi)/18, (13pi)/18, (17pi)/18, (25pi)/18, (29pi)/18, …`

 
`:.\ text(Solutions)`

`x = 4sin\ pi/18 \ \ \ (= 4sin\ (17pi)/18)`

`x = 4sin\ (5pi)/18 \ \ \ (= 4sin\ (13pi)/18)`

`x = 4sin\ (25pi)/18 \ \ \ (= 4sin\ (29pi)/18)`
 

`text(If roots of)\ \ x^3-12x + 8 = 0\ \ text(are)\ \ α, β, γ:`

`α + β + γ = -b/a = 0`

`αβ + βγ + αγ = c/a = -12`

`(4sin\ pi/18)^2 + (4sin\ (5pi)/18)^2 + (4sin\ (25pi)/18)^2` `= (α + β + γ)^2-2(αβ + βγ + αγ)`
`16(sin^2\ pi/18 + sin^2\ (5pi)/18 + sin^2\ (25pi)/18)` `= 0-2(-12)`
`:. sin^2\ pi/18 + sin^2\ (5pi)/18 + sin^2\ (25pi)/18` `= 24/16=3/2`

Functions, EXT1 F2 EQ-Bank 26

The polynomial  `P(x) = x^3 + px^2 + qx + r`  has roots `sqrtk, -sqrtk` and `alpha`.

  1.  Explain why  `alpha + p = 0`.   (1 mark)

  2.  Show that  `kalpha = r`.   (1 mark)

  3.  Show that  `pq = r`.   (2 marks)

Show Answers Only

a.    `text(See Worked Solutions)`

b.    `text(See Worked Solutions)`

c.    `text(See Worked Solutions)`

Show Worked Solution

a.    `text(Sum of roots:)`

`sqrtk-sqrtk + alpha` `= -b/a`
`alpha` `= -p`
`alpha + p` `= 0`

 

b.    `text(Product of roots:)`

`sqrtk xx -sqrtk xx alpha` `= -d/a`
`-kalpha` `= -r`
`:.kalpha` `= r`

 

c.     `sqrtk(-sqrtk) + sqrtk alpha-sqrtk alpha` `= c/a`
  `-k` `= q`

 
`text(Substitute)\ \ k = -q\ \ text(into part (b)):`

`-qalpha` `= r`
`-q xx -p` `= r\ \ \ text{(using part(a))}`
`:. pq` `= r`

Functions, EXT1 F2 EQ-Bank 16

The polynomial  `P(x) = x^3-2x^2 + kx + 24`  has roots  `alpha, beta, gamma`.

  1. Find the value of  `alpha + beta + gamma`.   (1 mark)

  2. Find the value of  `alphabetagamma`.   (1 mark)

  3. It is known that two of the roots are equal in magnitude but opposite in sign.

     

    Find the third root and hence find the value of `k`.   (2 marks)

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a.    `2`

b.    `−24`

c.    `−12`

Show Worked Solution

a.    `alpha + beta + gamma = −b/a = 2`
 

b.    `alphabetagamma = −d/a = −24`
 

c.    `text(Let roots be)\ \ alpha, −alpha, \ beta:`

`alpha-alpha + beta` `= 2`
`beta` `= 2`

 
`text(Substitute)\ \ beta = 2\ \ text(into equation:)`

`2^3-2 ·2^2 + 2k + 24` `= 0`
`2k` `= -24`
`k` `= -12`

Functions, EXT1 F2 EQ-Bank 20

The equation  `2x^3 + x^2-kx + 6 = 0`  has 2 roots which are reciprocals of each other.

Find the value of `k`.   (2 marks)

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`13`

Show Worked Solution

`text(Let roots be)\ \ alpha, \ 1/alpha, \ beta:`

COMMENT: Substituting  `beta=-3`  into the equation solves this question efficiently.

`alpha · 1/alpha · beta` `= -d/a`
`beta` `= -6/2=-3`

 
`text(Substitute)\ \ beta = -3\ \ text(into equation:)`

`2(-3)^3 + (-3)^2-(-3)k + 6` `= 0`
`-54 + 9 + 3k + 6` `= 0`
`3k` `= 39`
`k` `= 13`

Functions, EXT1 F2 2019 HSC 7 MC

Let  `P(x) = qx^3 + rx^2 + rx + q`  where `q` and `r` are constants, `q != 0`. One of the zeros of  `P(x)`  is  `-1`.

Given that  ` alpha`  is a zero of  `P(x),\ alpha != -1`, which of the following is also a zero?

A.     `-1/alpha`

B.     `-q/alpha`

C.     `1/alpha`

D.     `q/alpha`

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`C`

Show Worked Solution

`text(Roots:)\ \ alpha,\ beta,\ -1`

Mean mark 51%.

`alpha beta(-1) = -d/a = -q/q = -1`

`- alpha beta` `= -1`
`:. beta` `= 1/alpha`

 
`=>  C`

Functions, EXT1′ F2 2018 HSC 3 MC

The cubic equation  `x^3 + 2x^2 + 5x-1 = 0`  has roots, `alpha`, `beta` and `gamma`.

Which cubic equation has roots  `(−1)/alpha, (−1)/beta, (−1)/gamma`?

  1. `x^3-5x^2-2x + 1 = 0`
  2. `x^3-5x^2-2x-1 = 0`
  3. `x^3 + 5x^2 + 2x + 1 = 0`
  4. `x^3 + 5x^2-2x + 1 = 0`
Show Answers Only

`text(D)`

Show Worked Solution

`alpha beta gamma = 1`

`alpha beta + beta gamma + gamma alpha = 5`

`alpha + beta + gamma = −2`
 

`text(If roots are)\ \ (−1)/alpha, (−1)/beta, (−1)/gamma :`

`(−1)/alpha · (−1)/beta · (−1)/gamma = (−1)/(alphabetagamma) = −1`

`=> d = 1`
 

`1/(alphabeta) + 1/(betagamma) + 1/(gammaalpha) = ((alpha + beta + gamma))/(alphabetagamma) = −2`

`=>\ c = −2`
 

`(−1)/alpha-(−1)/beta-(−1)/gamma = (−(alphabeta + betagamma + gammaalpha))/(alphabetagamma) = −5`

`=> b = 5`
 

`:.\ text(Equation is)\ \ x^3 + 5x^2-2x + 1 = 0`

`=>D`

Functions, EXT1′ F2 2017 HSC 5 MC

The polynomial  `p(x) = x^3-2x + 2`  has roots `alpha`, `beta` and `gamma`.

What is the value of  `alpha^3 + beta^3 + gamma^3`?

  1. `−10`
  2. `−6`
  3. `−2`
  4. `0`
Show Answers Only

`B`

Show Worked Solution

`text(S)text(ince)\ \ alpha, beta, gamma\ text(are roots,)`

`p(alpha)` `= alpha^3-2alpha + 2 = 0` `…\ (1)`
`p(beta)` `= beta^3-2beta + 2 = 0` `…\ (2)`
`p(gamma)` `= gamma^3-2gamma + 2 = 0` `…\ (3)`

 
`text(Add:)\ (1) + (2) + (3)`

`alpha^3 + beta^3 + gamma^3-2(alpha + beta + gamma) + 6 = 0`

`:. alpha^3 + beta^3 + gamma^3=-6\ \ \ \ text{(note:}\ \ alpha + beta + gamma =- b/a=0 text{)}`

`=> B`

Functions, EXT1′ F2 2017 HSC 13b

Let `a, b` and `c` be real numbers. Suppose that  `P(x) = x^4 + ax^3 + bx^2 + cx + 1`  has roots  `alpha, 1/alpha, beta, 1/beta,`  where  `alpha > 0 and beta > 0`.

Prove that  `a = c`.   (2 marks)

 

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`text(Proof)\ \ text{(See Worked Solutions)}`

Show Worked Solution

`text(Sum of roots) = -b/a:`

`-a = alpha + 1/alpha + beta + 1/beta`
 

`text(Sum of products of 3 roots) = -d/a:`

`- c` `= alpha · 1/alpha · beta + alpha · 1/alpha · 1/beta + alpha · beta · 1/beta + 1/alpha · beta · 1/beta`
  `= beta + 1/beta + alpha + 1/alpha`
  `= -a`

 
`:. a = c\ text(… as required.)`

Functions, EXT1′ F2 2012 HSC 5 MC

The equation  `2x^3 − 3x^2 − 5x − 1 = 0`  has roots  `α`, `β` and `γ`.

What is the value of  `1/(α^3β^3γ^3)`?

  1. `1/8`
  2. `−1/8`
  3. `8`
  4. `−8` 
Show Answers Only

`C`

Show Worked Solution
`αβγ` `=(-d)/a= 1/2`
`:.α^3β^3γ^3`  `=(1/2)^3=1/8`
`:.1/(α^3β^3γ^3)`  `=8`

 
`=>C`

Functions, EXT1′ F2 2007 HSC 3b

The zeros of  `x^3-5x + 3`  are `alpha, beta` and `gamma.`

Find a cubic polynomial with integer coefficients whose zeros are `2 alpha, 2 beta` and `2 gamma.`   (2 marks)

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`x^3-20x + 24`

Show Worked Solution

`text(Solution 1)`

`\alpha +\beta+\gamma` `=0`  
`2\alpha +2\beta+2\gamma` `=2(\alpha +\beta+\gamma)=0`  

 

`\alpha\beta +\beta\gamma+\gamma\alpha` `=-5`  
`2\alpha2\beta +2\beta2\gamma+2\gamma2\alpha` `=4(\alpha\beta +\beta\gamma+\gamma\alpha)=-20`  

 

`\alpha\beta\gamma` `=-3`
`2\alpha2\beta2\gamma` `=8\alpha\beta\gamma=-24`

 
`:.\ text(Polynomial is)\ \ x^3-20x + 24`
 

`text(Solution 2)`

`P(x) = x^3-5x + 3`

`text(New zeros are)\ 2 alpha, 2 beta and 2 gamma:`

`H(x)` `=(x/2)^3-5(x/2) + 3`
  `=x^3/8-(5x)/2+3`

 
`:.\ text(Polynomial with integer coefficients is)\ \ x^3-20x + 24`

Functions, EXT1 F2 2006 HSC 4a

The cubic polynomial  `P(x) = x^3 + rx^2 + sx + t`. where `r, \ s` and `t` are real numbers, has three real zeros,  `1, alpha` and `-alpha.`

  1. Find the value of  `r.`   (1 mark)

  2. Find the value of  `s + t.`   (2 marks)

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a.    `-1`

b.    `0`

Show Worked Solution

a.    `P(x) = x^3 + rx^2 + sx + t`

`text(Roots are)\ \ 1, alpha, -alpha`

`1 + alpha + -alpha` `= – b/a`
`1` `= – r/1`
`:.\ r` `= -1`

 

b.    `P(x) = x^3-x^2 + sx + t`

`P(1) = 0`

`0` `= 1-1 + (s xx 1) + t`
`0` `= s + t`

Functions, EXT1 F2 2008 HSC 2c

The polynomial `p(x)` is given by  `p(x) = ax^3 + 16x^2 + cx-120`, where `a` and `c` are constants.

The three zeros of `p(x)` are `-2,  3` and `beta`.

Find the value of `beta`.   (3 marks) 

Show Answers Only

`-5`

Show Worked Solution

`p(x) = ax^3 + 16x^2 + cx-120`

`text(Roots:)\ \ -2, \ 3, \ beta`

`-2 + 3 + beta` `= -B/A`
`beta + 1` `= -16/a`
`beta` `= -16/a-1\ \ \ \ \ …\ (1)`

 

`-2 xx 3 xx beta` `= -D/A`
`-6 beta` `= 120/a`
`beta` `= -20/a\ \ \ \ \ …\ (2)`

 

MARKER’S COMMENT: Many students displayed significant inefficiencies in solving simultaneous equations.
`- 16/a-1` `= -20/beta`
`-16-a` `= -20`
`a` `= 4`

 
`text(Substitute)\ \ a = 4\ \ text(into)\ (1):`

`beta=-16/4-1= -5`

Functions, EXT1 F2 2014 HSC 5 MC

Which group of three numbers could be the roots of the polynomial equation  `x^3 + ax^2 − 41x + 42 = 0`?

  1. `2, 3, 7`
  2. `1, −6, 7`
  3. `−1, −2, 21`
  4. `−1, −3, −14`
Show Answers Only

`B`

Show Worked Solution

`text(Let roots be)\ alpha, beta, gamma`

`alpha beta gamma = -d/a = -42`

`:.\ text(Cannot be)\ A\ text(or)\ C`

`alpha beta + beta gamma + gamma alpha = c/a = -41`

`:.\ text(Cannot be D)`

`=>  B`

Functions, EXT1 F2 2012 HSC 3 MC

A polynomial equation has roots  `alpha`,  `beta`,  `gamma`  where

`alpha + beta + gamma = -2`,    `alphabeta + alphagamma + betagamma = 3`,    `alphabetagamma = 1`.

Which polynomial equation has the roots  `alpha`,  `beta`, and  `gamma`?

  1. `x^3 + 2x^2 + 3x + 1 = 0`
  2. `x^3 + 2x^2 + 3x- 1 = 0`
  3. `x^3- 2x^2 + 3x + 1 = 0`
  4. `x^3- 2x^2 + 3x- 1 = 0`
Show Answers Only

`B`

Show Worked Solution

`text(Using)\ \ ax^3 + bx^2 + cx + d = 0`

`text(By elimination)`

`alpha beta gamma = -d/a  = 1`

`:.\ text(Cannot be)\ A\ text(or)\ C\ \ (text(where)\ alphabetagamma = -1 text{)}`

`alpha + beta + gamma = -b/a = -2`

`:.\ text(Cannot be)\ D\ \ (text(where)\ alpha + beta + gamma = 2 text{)}`

`=> B`