smc-4242-10-Factor Theorem

Polynomials, SMB-012

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

Show `h(1)=0` (1 mark)

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

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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

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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`.

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

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

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`text(See Worked Solutions)`
`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`.

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

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

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`text(See Worked Solution)`
`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`.

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

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Find the other zeros. (2 marks)

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`text(See Worked Solution)`
`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-003 MC

If `x + a` is a factor of `8x^3-14x^2-a^2 x`, then the value of `a` is

7
4
1
–2

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`D`

Show Worked Solution

`f(-a)`	`= 8(-a)^3-14(-a)^2-a^2(-a)`
`0`	`= -8a^3-14a^2 + a^3`
`0`	`= -7a^3-14a^2`
`0`	`= -7a^2 (a + 2)`
`a`	`= -2`

`=>D`

Functions, EXT1 F2 2020 HSC 11a

Let `P(x) = x^3 + 3x^2-13x + 6`.

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

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

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`text(See Worked Solutions)`
`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`.

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

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Find the other zeros. (2 marks)

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`text(See Worked Solution)`
`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 2016 HSC 2 MC

What is the remainder when `2x^3-10x^2 + 6x + 2` is divided by `x-2`?

`-66`
`-10`
`-x^3 + 5x^2-3x-1`
`x^3-5x^2 + 3x + 1`

Show Answers Only

`B`

Show Worked Solution

`P(2)`	`= 2 · 2^3-10 · 2^2 + 6 · 2 + 2`
	`= -10`

`=> B`

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`.

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

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

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`text(See Worked Solutions)`
`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 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)

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`-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`