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Calculus, EXT1 EQ-Bank 28

\(P(x)\) is a polynomial where  \(P(\alpha)=0\)  and  \(P^{\prime}(\alpha)=0\).

  1. Show that \((x-\alpha)^2\) is a factor of \(P(x)\).   (2 marks)

  2. The curve  \(y=x^3+b x^2+c x+4\)  is tangent to the \(x\)-axis at  \(x=-1\). Find the values of \(b\) and \(c\).   (3 marks)

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

\(P(x)=(x-\alpha) \cdot Q(x)\)
 

\(P^{\prime}(x)=Q(x)+(x-a) \cdot Q(x)\)

\(\text{Since}\ \ P^{\prime}(\alpha)=0:\)

\(Q(\alpha)=0 \ \ \Rightarrow\ \ (x-\alpha) \ \text{is a factor of} \ \ Q(\alpha)\)
 

\(\text{Let} \ \ Q(x)=(x-\alpha) \cdot R(x)\)

\(P(x)=(x-\alpha)^2 \cdot R(x)\)

\(\therefore \ (x-\alpha)^2 \ \text{is a factor of} \ P(x).\)
 

b.    \(b=6, c=9\)

Show Worked Solution

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

\(P(x)=(x-\alpha) \cdot Q(x)\)
 

\(P^{\prime}(x)=Q(x)+(x-a) \cdot Q(x)\)

\(\text{Since}\ \ P^{\prime}(\alpha)=0:\)

\(Q(\alpha)=0 \ \ \Rightarrow\ \ (x-\alpha) \ \text{is a factor of} \ \ Q(\alpha)\)
 

\(\text{Let} \ \ Q(x)=(x-\alpha) \cdot R(x)\)

\(P(x)=(x-\alpha)^2 \cdot R(x)\)

\(\therefore \ (x-\alpha)^2 \ \text{is a factor of} \ P(x).\)
 

b.    \(\text{Since the curve is tangent at} \ \ x=-1\)

\(x=-1 \ \ \text{is a double root}\)

\(P(x)=x^3+b x^2+c x+4\)

\(P(-1)=-1+b-c+4=0 \ \ \Rightarrow\ \ b-c=-3\ \ldots\ (1)\)
 

\(P^{\prime}(x)=3 x^2+26 x+c\)

\(P^{\prime}(-1)=3-2 b+c=0 \ \ \Rightarrow\ \ -2 b+c=-3\ \ldots\ (2)\)
 

\(\text{Add} \ (1)+(2):\)

\(-b=-6 \ \ \Rightarrow\ \ b=6\)

\(\text{Substitute \(\ b=6\ \) into (1):}\)

\(-6-c=-3 \ \ \Rightarrow\ \ c=9\).

\(\therefore b=6, c=9\)

Functions, EXT1′ F2 2016 HSC 13d

Suppose  `p(x) = ax^3 + bx^2 + cx + d`  with `a, b, c` and `d` real, `a != 0.`

  1. Deduce that if  `b^2-3ac < 0`  then `p(x)` cuts the `x`-axis only once.   (2 marks)

  2. If  `b^2-3ac = 0`  and  `p(-b/(3a)) = 0`, what is the multiplicity of the root  `x = -b/(3a)?`   (2 marks)

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a.    `text(See Worked Solutions)`

b.    `3`

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a.    `p(x) = ax^3 + bx^2 + cx + d`

`p^{′}(x) = 3ax^2 + 2bx + c`

`=> p(x)\ text(will cut the)\ xtext(-axis once only if)\ \ Delta(p^{′}(x)) < 0`

`(2b)^2-4(3a)c` `< 0`
`4b^2-12ac` `< 0`
`b^2-3ac` `< 0`

 

b.    `p(-b/(3a)) = 0`

`p^{′}(−b/(3a))` `=3a(- b/(3a))^2 + 2b (- b/(3a))+c`
  `=- b^2/(3a)+c`
  `=0\ \ \ (text{given}\ \ b^2-3ac = 0)`

 
`:.\ text(Multiplicity at least 2.)`

♦♦ Mean mark 32%.

 
`p^{″}(x) = 6ax + 2b`

`p^{″}(−b/(3a))=6a(−b/(3a)) + 2b=0`

`:. text(Multiplicity of)\ \ x=− b/(3a)\ \ text(is 3.)`

Functions, EXT1′ F2 2016 HSC 2 MC

Which polynomial has a multiple root at  `x = 1?`

  1. `x^5-x^4-x^2 + 1`
  2. `x^5-x^4-x-1`
  3. `x^5-x^3-x^2 + 1`
  4. `x^5-x^3-x + 1`
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`=> C`

Show Worked Solution

`\text{By trial and error}`

`text(Consider option)\ C:`

`P(1)` `= 1-1-1 + 1 = 0`
`P^{′}(x)` `= 5x^4-3x^2-2x`
`P^{′}(1)` `= 5-3-2 = 0`

 
`:.\ text(Multiple root at)\ x = 1`

`=> C`

Functions, EXT1′ F2 2014 HSC 14a

Let  `P(x) =x^5-10x^2 +15x-6`.

Show that  `x = 1`  is a root of `P(x)` of multiplicity three.   (2 marks)

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`text{See Worked Solutions}`

Show Worked Solution

`P(x) =x^5-10x^2 +15x-6`

`P(1) = 1-10 + 15-6 = 0`
 

`P^{′}(x)` `= 5x^4-20x + 15`
`P^{′}(1)` `= 5-20 + 15 = 0`
`P^{″}(x)` `= 20x^3-20`
`P^{″}(1)` `= 20-20 = 0`
`P^{‴}(x)` `= 60x^2`
`P^{‴}(1)` `= 60 ≠ 0`

 
`:.x = 1\ text(is a root of)\ P(x)\ text(of multiplicity 3.)`