polynomial

POLYNOMIAL XI SCIENCE

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A. The General Aspects of Polynomial 1. Definition and Components

Algebraic forms, such as a linear equation and quadratics equation are kinds of

polynomial equation of degree 1 and 2. Consider the following algebraic forms :

a. x2 – 3x + 5

b. 2x3 + 7x

2 – 7x + 2

c. 4x5 – 5x

3 – 2x

2 – 7

The algebraic forms above are called polynomial in variable x that accomadate in the

polynomial. They consecutively have degree of 2, 3 and 5.

Definition :

The general form of polynomial I variable x of degree n is :

P(x) = an xn + an–1 x

n–1 + an–2 x

n–2 + an–3 x

n–3 + … + a1 x + a0

The polynomial are arraged based on descending order of the exponent of x, where :

an, an–1, …, a0 = real coefficients of polynomial and an ≠ 0,

a0 = real constant

n = degree of polynomial, n is whole number.

2. Algebraic Operation of Polynomial

Algebraic equation of polynomial such as addition, subtraction, multiplication and

division have similar characteristics and rules as the algebraic operation on real numbers.

Given f(x) = x3 + 7x

2 + 3 and g(x) = x

5 + 3x

4 – 7x

2 – 3x + 4.

Determine : a. f(x) + g(x) and its degree

b. f(x) – g(x) and its degree

c. f(x) × g(x) and its degree

Answer :

a. f(x) + g(x) = x3 + 7x

2 + 3 + x

5 + 3x

4 – 7x

2 – 3x + 4

= x5 + 3x

4 + x

3 – 3x + 7

and has degree 5.

b. f(x) – g(x) = x3 + 7x

2 + 3 – (x

5 + 3x

4 – 7x

2 – 3x + 4)

= x3 + 7x

2 + 3 – x

5 – 3x

4 + 7x

2 + 3x – 4

= –x5 – 3x

4 + 14x

2 + 3x – 1

and has degree 5.

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c. f(x) × g(x) = (x3 + 7x

2 + 3) (x

5 + 3x

4 – 7x

2 – 3x + 4)

= x8 + 3x

7 – 7x

5 – 3x

4 + 4x

3 + 7x

7 + 21x

6 – 49x

4 – 21x

3 + 28x

2 + 3x

5 + 9x

4 –

21x2 – 9x + 12

= x8 + 10x

7 + 21x

6– 5x

5 – 43x

4 – 17x

3 + 7x

2 – 9x + 12

and has degree 8.

3. Polynomial Equation

Definition

Suppose given polynomial f(x) and g(x), where

f(x) = an xn + an–1 x

n–1 + an–2 x

n–2 + an–3 x

n–3 + … + a1 x + a0

g(x) = bn xn + bn–1 x

n–1 +bn–2 x

n–2 + bn–3 x

n–3 + … + b1 x + b0

f(x) = g(x) if satisfies :

an = bn , an–1 = bn–1 , … , a1 = b1 , a0 = b0.

Find the value of a and b of equation x3 + 4x

2 – 7x + a = (x – 2) (x + 1) (x + b) !

Answer :

x3 + 4x

2 – 7x + a = (x – 2) (x + 1) (x + b)

= (x2 – x – 2) (x + b)

x3 + 4x

2 – 7x + a = x

3 + (b – 1)x

2 – (b + 2)x – 2b

4 = b – 1 → coefficient of x2

b = 5

a = –2b → constant

a = –2 . 5

a = –10

Thus, the value of a = –10 and b = 5.

1. Given polynomial of f(x) = –7x5 + x

4 + 5x

2 – 3. Determine :

a. The degree of the polynomial

b. The coefficient of x4

c. The coefficient of x3

d. The constant

2. Among the algebraic forms below which one is a polynomial and which one is not

polynomial!

a. (3x2 – 2) (x

2 + 3x – 1) c. (2x + 5) 42 −x

b. (2x – 1) (x + x

1) d.

2log (x

2 – 3x + 2)

3. Given the polynomial of f(x) = 4 – x2, g(x) = x + 2 and h(x) = x

3 – 3x

2 + 3. Determine :

a. f(x) + g(x) and its degree

b. f(x) + g(x) – h(x) and its degree

c. f(x) × g(x) and its degree

d. f(x) – g(x) – h(x) and its degree


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4. Find the coefficient of p and q on the following equations !

a. (x2 + 1) (x + p) – 12 = (x – 1) (x + 2) (x – 3) + 10x

2 – 10x – 2

b. 65

135

23 2 +−

−=

−+

− xx

x

x

q

x

p

5. Determine the value of d, e and f from the following equation!

6116

18226

321 23

2

++−

+−=

−+

−+

− xxx

xx

x

f

x

e

x

d

B. The Value of a Polynomial 1. Determining the value of polynomial by substitution method

Suppose a polynomial of f(x) = ax3 + bx

2 + cx + d. if the value of x replaced by k, then

the value of polynomial of f(x) for x = k is f(k) = ak3 + bk

2 + ck + d.

Determine the value of polynomial p(x) = x4 – 3x

3 – 2x + 3 for x = 1 and x = –2 !

Answer :

For x = 1 → p(1) = 14 – 3 . 1

3 – 2 . 1 + 3

= 1 – 3 – 2 + 3

= –1

For x = –2 → p(–2) = (–2)4 – 3 . (–2)

3 – 2 . (–2) + 3

= 16 + 24 + 4 + 3

= 47

2. Determining the value of polynomial by Scheme (or Horner Method)

The first step of using scheme method (Horner Method) is to write a polynomial from

the right to left starting gfform the highest exponent of the variable. To solve example 5.3 with

p(x) = x4 – 3x

3 – 2x + 3 for x = 1, you must write each coefficient each term of the polynomial

in the chart. Consider the following chart

p(x) = x4 – 3x

3 + 0x

2 – 2x + 3

Thus, the value of p(1) = –1

1. Calculate the value of the following polynomial using the subsitution method!

a. f(x) = x3 + 7x

2 – 4x + 3, for x = 5

b. g(x) = 2x3 + 4x

2 + 6x + 8, for x = 3

c. h(x) = 2x3 + 4x

2 – 18, for x = 3

d. p(x) = 5x4 + 7x

2 + 3x + 1, for x = –1

e. s(x) = x3 – x + 1, for x =

3

1−

1 –3 0 –2 3

1 –2 –2 –4

x = 1

1 –2 –2 –4 –1

+


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2. Calculate the value of the following polynomial using the Horner method!.

a. p(x) = x3 + 7x

2 – 2x + 4, for x = 2

b. f(x) = 2x4 – x

2 + 8, for x = –3

c. g(x) = 7x4 + 20x

3 – 5x

2 + 3x + 5, for x = 1

d. h(x) = 4x7 – 8x

5 + 4x

4 – 5x

3 + 15x – 22, for x = –2

e. k(x) = x5 + x

4 – 2x

3 + 2x – 1, for x = –1

3. Determine the value of z so that the polynomial

a. x3 – (2z – 1)x

2 + 7x – 3 is 0 for x = –1

b. 2x4 –

−

3

1zx – 5x + 7 is 73 for x = 1

C. The Polynomial Dividing 1. Definition

The procedure of dividing a polynomial using long division is similar to the procedure of

dividing integers.

Determine the quotient and remainder of divison of P(x) = 2x3 + 4x

2 – 18 by x – 3 !

Answer :

2x2 + 10x + 30

x – 3 2x3 + 4x

2 + 0x – 18

2x3 – 6x

2

10x2 + 0x – 18

10x2 – 30x

30x – 18

30x – 90

72

Using the method, we get the quotient H(x) = 2x2 + 10x + 30 and the remainder S(x) = 72.

Consider the following expression

P(x) = Q(x) . H(x) + S(x)

The expression above is division of polynomial, where

P(x) = the dividend, Q(x) = the divisor, H(x) = the quotient and

S(x) = the remainder

the quotient the dividend

the remainder

the divisor

The Long Division


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The Horner Method is short way to solve the division of polynomial. This method

can solve several form of division polynomial.

2. Dividing Polynomial by (x – k)

The principle of Horner method s derived from the concept of long division. To

understand it, consider the division of polynomial of P(x) = ax2 + bx + c by (x – k) below

Consider that the remainder, namely ak2 + bk + c is a P(x) for x = k. This fact shows

that the remainder of te polynomial division for P(x) by (x – k) can be determined by

calculating the value of the polynomial on x = k. The condition of x = k is yielded if the

divisor (x – k) valued at zero. Next, inspect the determining of the value of polynomial P(x) =

ax2 + bx + c for x = k using the scheme that you have learned.

Determine the quotient and the remainder of P(x) = 3x3 – 4x

2 + x + 7 divided by (x – 2) using

Horner method!

Answer :

So the quotient is H(x) = 3x2 + 2x + 5 and the remainder is S(x) = 17

HORNER METHOD

ax + (b + ak)

(x – k) ax2 + bx + c

ax2 – akx

(b + ak)x + c

(b + ak)x – bk – ak2

ak2 + bk + c

3x3 – 4x

2 + x + 7

x = 2 3 –4 1 7

6 4 10

3 2 5 17

+

the remainder

the coefficient quotient

ax2 + bx + c

x = k a b c

ak bk + ak2

a (b + ak) ak2 + bk + c

+ the remainder



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3. Dividing Polynomial by (ax – b)

The division of polynomial by (ax – b) is expansion of divison of the polynomial P(x)

by (x – k). The general form of dividing a polynomial by (ax – b) with the quotient H(x) and

the remainder S(x) is :

The polynomial form of dividing a polynomial by (x – k) with the quotient H(x) and

the remainder of S(x) is P(x) = (x – k) . H(x) + S(x)

Suppose k = a

b then get

P(x) =

−a

bx . H(x) + S(x) =

a

bax )( − . H(x) + S(x) = (ax – b) .

a

x)(H + S(x)

Based on the fact above we have the following things.

� The polynomial of P(x) divided by (ax – b) is resulting a

x)(H, which H(x) is quotient of

dividing of P(x) by

−a

bx .

� The remainder of dividing of P(x) divided by (ax – b) is the same as the remainder of

dividing it by

−a

bx .

Determine the quotient and the remainder of P(x) = 3x4 – 2x

2 + x – 3 divided by (3x + 1) using

Horner method!

Answer :

The divisor will zero if x = 3

1−

So the quotient is H(x) = 3

9

14

3

53 23 +−− xxx

= 27

14

9

5

3

1 23 +−− xxx and S(x) = 27

95− .

P(x) = (ax – b) . H(x) + S(x)

3x4 + 0x

3 – 2x

2 + x – 3

x =3

1− 3 0 –2 1 –3

–1 3

1

9

5

27

14−

3 –1 3

5−

9

14

27

95−

+

the remainder



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4. Dividing Polynomial by (ax2 + bx + c)

The division of polynomial by (ax2 + bx + c) using the Horner method can be done if

the divisor (ax2 + bx + c) can be factorized. The general form of division is :

With H(x) and S(x) are consecutively the quotient and the remainder of the division.

Suppose (ax2 + bx + c) can be factorized to P1 . P2 therefore ax

2 + bx + c = P1 . P2, then :

The step to determine the quotient and the remainder of the division of a polynomial are :

� Divide the polynomial of P(x) by P1 to get H1(x) and S1 as the quotient and the remainder

� Divide H1(x) by P2 to get H(x) and S2 as quotient and the remainder

� The quotient of dividing P(x) by (Q(x) = P1 . P2) is H(x), and the remainder is

S(x) = P1 . S2 + S1.

Determine the quotient and the remainder of 2x4 + 3x

3 – x + 5 divided by x

2 + x – 2!

Answer :

P(x) = 2x4 + 3x

3 – x + 5

Q(x) = x2 + x – 2 = (x – 1) (x + 2) ⇒P1 = (x – 1), valued zero for x = 1

P2 = (x + 2), valued zero for x = –2

Then use the following Horner method.

Thus, the quotient of dividing 2x4 + 3x

3 – x + 5 by x

2 + x – 2 is H(x) = 2x

2 + x + 3, while the

remainder can be calculated by formula of S(x) = P1 . S2 + S1

S(x) = P1 . S2 + S1

= (x – 1) . (–2) + 9

= –2x + 2 + 9

= –2x + 11.

P(x) = (ax2 + bx + c) . H(x) + S(x)

P(x) = (ax2 + bx + c) . H(x) + S(x) = P1 . P2 . H(x) + S(x)

P(x) = 2x4 + 3x

3 + 0x

2 – x + 5

x = 1 2 3 0 –1 5

2 5 5 4

2 5 5 4 9

+

= S1

H1(x) = 2x3 + 5x

2 + 5x + 4

x = –2 2 5 5 4

–4 –2 –6

2 1 3 –2

+

= S2


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1. Determine the quotient and the remainder of the following polynomial division:

a. P(x) = x3 + 2x

2 + 3x + 6 divided by (x – 2)

b. P(x) = x3 + 4x

2 + x + 3 divided by (2x – 1)

c. P(x) = 3x3 + 4x

2 – 7x + 1 divided by (x + 3)

d. P(x) = x4 – x

2 + 7 divided by (x

2 – 3x + 2)

e. P(x) = x4 + 6x

2 + 3x – 15 divided by (2x

2 + 3x – 2)

f. P(x) = 2x5 – 4x

3 – 5x + 9 divided by (x

2 + 1)

2. Determine the value of k and the quotient of the following division!

a. P(x) = 5x4 + 3x

3 + x

2 + kx – 8 divided by (x + 2), with the remainder = 42

b. P(x) = 3x5 + kx – 6 divided by (x – 2), with the remainder = 2

c. P(x) = x4 + kx

3 – 2x

2 + 4x – 1 divided by (2x – 4), with the remainder = 47

d. P(x) = 6x3 – x

2 + kx + 1 divided by (3x + 1), with the remainder = 0

3. Determine the value of m + n and the quotient of the following division :

a. P(x) = x4 – 4x

3 + 7x

2 + mx + n , divided by (x

2 – 2x – 3), with the remainder = 8x + 10

b. P(x) = 4x4 – 8x

3 + mx

2 – nx – 1, divided by (2x

2 + 3x – 1), with the remainder = 2x + 2

More exercise :

http://edhelper.com/polynomials225.htm



5. The Remainder Theorem with divisor (x – k)

Polynomial in variable x, namely P(x), can also be considered as fuction on x, namely

f(x). The notation f(x) will also be used to represent polynomial.

Determine the value of z so that the remainder of f(x) = 2x3 – 3x

2 + zx – 1 divided by (x – 2) is

9!

Answer :

The remainder is value f(x) for x = 2

1. By substituting

S = f(2)

9 = 2.23 – 3.2

2 + z.2 – 1

9 = 16 – 12 + 2z – 1

6 = 2z

z = 3

Theorem:

If polynomial of f(x) with degree of n divided by (x – k), then the remainder is S(x) = f(k)


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2. By Horner method

2z + 3 = 9

z = 3

6. The Remainder Theorem with divisor (ax – k)

Determine the remainder of f(x) = 6x3 – 2x

2 – x + 7 divided by (3x + 2)!

Answer :

The remainder is value f(x) for x = 3

2−

1. By substituting

S =

−3

2f

S = 73

2

3

2.2

3

2.6

23

+

−−

−−

−

S = 73

2

9

8

9

16++−−

S = 5

2. By Horner method

S = 5

2x3 – 3x

2 + zx – 1

x = 2 2 –3 z –1

4 2 2z + 4

2 1 z + 2 2z + 3

+ = S

Theorem:

If polynomial of f(x) with degree of n divided by (ax – k), then the remainder is S(x) =

a

kf

6x3 – 2x

2 – x + 7

x = 3

2− 6 –2 –1 7

–4 4 –2

6 –6 3 5

+


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7. The Remainder Theorem with divisor (x – a) ( x – b)

Dividing polynomial of f(x) by (x – a) (x – b) with the quotient of H(x) and the

remainder of S(x), is written as :

Since the degree of the divisor is 2, the degree of S is at most 1. Let S = (px + q), then the

division can written as follows.

If polynomial of f(x) divided by (x – 3) the remainder is 10, while if it is divided by (x + 4)

the remainder is 3, determine the remainder if f(x) is divided by (x2 + x – 12)!

Answer :

f(x) : (x – 3) then the remainder = 10 → f(3) = 10

f(x) : (x + 4) then the remainder = 3 → f(–4) = 3

Let S = px + q

f(x) = (x – 3) . (x + 4) . H(x) + (px + q)

for x = 3 → f(3) = 10 → 3p + q = 10

for x = –4→ f(–4) = 3 → –4p + q = 3 –

7p = 7

p = 1

3p + q = 10

3.1 + q = 10

q = 7

So, the remainder of S(x) = px + q = x + 7

1. Using reminder theorem, determine the remainder of the following division :

a. x4 – 3x

3 + 5x

2 – 4x + 1 divided by x + 1

b. 2x3 – 4x

2 + 8x – 1 divided by x + 3

c. 8x4 – 2x

3 + 5x – 6 divided by 2x – 1

d. 6x4 – 3x

2 + 3x + 6 divided by x

2 + 2x – 3

2. Determine the value of p, q and r , given as follows

a. The polynomial 3x3 – px

2 – 13x + 10 is divisible by (x – 2)

b. The polynomial px3 – qx

2 + 5x + r if it is divided by (x – 3) the remainder is 14 and the

remainder is 46 when divided by (x – 2).

c. The polynomial 12x3 – px

2 – qx – 1 is divisible by (3x + 1) and provide the remainder

63 when divided by (x – 2)

3. The polynomial of f(x) divided (2x + 1) the remainder is 3 and if divided (x – 2) the

remainder is 18. Determine the remainder if f(x) divided by (2x2 – 3x – 2)!

f(x) = (x – a) (x – b) . H(x) + S(x)

f(x) = (x – a) (x – b) . H(x) + (px + q)


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4. The polynomial of f(x) divided by (x2 – 4) the remainder is (x + 2) and divided by (x – 3)

the remainder is 5. Determine the remainder if f(x) divided by (x2 – 5x + 6)!

5. The polynomial of f(x) divided by (x2 – 2x) the remainder is (5x + 1) and divided by (x

2 +

3x) the remainder is (3x + 1). Determine the remainder if f(x) divided by (x2 + x – 6)!

8. The Factorization Theorem

Show that (x – 4) is a factor of f(x) = 2x4 – 9x

3 + 5x

2 – 3x – 4.

Answer :

Using method of substitution, it will be:\

x – 4 = 0 → x = 4

f(x) = 2x4 – 9x

3 + 5x

2 – 3x – 4

f(4) = 2.44 – 9.4

3 + 5.4

2 – 3.4 – 4

= 512 – 576 + 80 – 12 – 4

= 0

Since f(4), then (x – 4) is a factor of f(x) = 2x4 – 9x

3 + 5x

2 – 3x – 4.

Determine the factors of 2x3 – 11x

2 + 17x – 6!

Answer :

The factor of –6 is ±1, ±2, ±3, ±6 and the factor of 2 is ±1, ±2, then the roots might be ±1,

±2

1, ±2, ±3, ±

2

3, ±6.

Checking each value to polynomial, we conclude

x = 1 → f(1) = 2.13 – 11.1

2 + 17.1 – 6 = 2 ≠ 0 → (x – 1) is not a factor

x = 2 → f(1) = 2.23 – 11.2

2 + 17.2 – 6 = 0 → (x – 2) is a factor

next, we use Horner method to determine other factor

f(x) = (x – 2) (2x2 – 7x + 3)

= (x – 2) (2x – 1) (x – 3)

Thus, the factor of 2x3 – 11x

2 + 17x – 6 is (x – 2), (2x – 1) and (x – 3).

Theorem:

(x – k) is factor from polynomial of f(x) if only if f(k) = 0

2x3 – 11x

2 + 17x – 6

x = 2 2 –11 17 –6

4 –14 6

2 –7 3 0

+


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1. Prove that –2 the root of x3 – 4x

2 – 4x + 16 = 0, then determine the other roots!

2. Prove that (4x – 1) is factor of 8x3 – 6x

2 + 9x – 2, then determine the other factors!

3. Prove that x + b is factor of x6 – b

6!

4. Determine the value of a and b, if

a. x2 + x – 6 is factor of 2x

3 + ax

2 + bx + 6

b. x2 + 5x + 6 is factor of x

3 + ax

2 + 31x + b

c. x2 – 3x + 2 is factor of ax

3 – 10x

2 + bx – 6

5. Determine the value of m so that the fraction of 65

622

2

−−

−+

xx

mxx can be simplified!

More exercise:








D. The Roots of a Polynomial Equation 1. Definition of the roots of Polynomial Equation

If 2 is the root of the equation 2x3 – x

2 – 5x – 2, determine the other roots!

Answer :

The quotient of the division of f(x) by x – 2 is 2x2 + 3x + 1, then

f(x) = (x – 2) . (2x2 + 3x + 1) = (x – 2) . (2x + 1) (x + 1)

Thus, the roots of f(x) are x = 2, x = 2

1− and x = –1.

2x3 – x

2 – 5x – 2

x = 2 2 –1 –5 –2

4 6 2

2 3 1 0

+

Theorem:

(x = k) is factor from polynomial of f(x) if only if f(k) = 0


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2. Rational Roots a Polynomial

a. If x1, x2 and x3 are the roots of ax3 + bx

2 + cx + d = 0

i. x1 + x2 + x3 = a

b−

ii. x1 . x2 + x1 . x3 + x2 . x3 = a

c

iii. x1 . x2 . x3 = a

d−

b. If x1, x2 , x3 and x4 are the roots of ax4 + bx

3 + cx

2 + dx + e = 0

i. x1 + x2 + x3 + x4 = a

b−

ii. x1 . x2 + x1 . x3 + x1 . x4 + x2 . x3 + x2 . x4 + x3 . x4 = a

c

iii. x1 . x2 . x3 + x1 . x2 . x4 + x1 . x3 . x4 + x2 . x3 . x4 = a

d−

iv. x1 . x2 . x3 . x4 = a

e

If one of the roots of the equation px3 – 5x

2 – 6x + 8 = 0 is 1, calculate the value of

a. x1 + x2 + x3

b. x1 . x2 + x1 . x3 + x2 . x3

c. x1 . x2 . x3

Answer :

The equation px3 – 5x

2 – 6x + 8 = 0 has three roots, suppose x1, x2 and x3. One of the roots is 1,

suppose x1. By substituting x = 1 into equation, we can calculate the coefficient of p

p.13 – 5.1

2 – 6.1 + 8 = 0

p – 5 – 6 + 8 = 0

p = 3

thus, the equation becomes 3x3 – 5x

2 – 6x + 8 = 0 → a = 3, b = –5, c = –6 and d = 8

a. x1 + x2 + x3 = a

b− =

3

5−− =

3

5

b. x1 . x2 + x1 . x3 + x2 . x3 = a

c =

3

6− = –2

c. x1 . x2 . x3 = a

d− =

3

8−

1. Determine the roots of following equation!

a. x3 – 4x

2 + 5x – 2 = 0

b. 2x3 – x

2 – 8x + 4 = 0

c. 6x4 – x

3 – 17x

2 + 16x – 4 = 0

d. 3x4 – 7x

3 + 5x

2 – 7x + 2 = 0


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2. Known one the roots of the equation 6x4 – 17x

3 – x

2 + kx – 5 = 0 is –1. Determine the value of

k and other roots!

3. The roots of mx3 – 14x

2 + 17x – 6 = 0 is x1, x2 and x3. For x1 = 3, determine the value of

a. x1 + x2 + x3

b. x1 . x2 + x1 . x3 + x2 . x3

c. x1 . x2 . x3

4. The equation of 2x3 – x

2 + nx + 4 = 0 has a pair of inverse roots.

a. Determine the value of n!

b. Determine the pair of roots!

polynomial

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