Polynomial long division & cubic equations

Polynomial One polynomial may be divided by another of lower degree by long division (similar long division to arithmetic long division).

Example (x3+ 3 x 2 + x + 9)

(x + 2)

Write the question in long division x+2 x3 + 3 x 2 + x + 9 form.

Begin with the x 3 term. 3 2 x divided by x equals x . x2 Place x 2 above the division bracket x+ 2 x3 + 3 xx 2 ++ 9 as shown.

subtract x3+ 2 x 2 Multiply x + 2 by x2 . 2 xx2( +2) = x 3 + 2 x 2 . x

3 2 3 2 Place x+ 2 x below x+ 3 x then subtract. x322+3 x −( x + 2 xx) = 2 x2 x+2 x3 + 3 xx 2 ++ 9

3 2 “Bring down” the next term (x) as x+2 x ↓ indicated by the arrow. x2 + x

Repeat this process with the result x2 + x − 1 until there are no terms remaining. x+2 x3 + 3 xx 2 ++ 9 2 x divided by x equals x. x3+2 x 2 ↓ ↓ 2 xx( +2) = x + 2 x . 2 x+ x ↓ 2 2 + − + =− 2 ( xx) ( x2 x) x x+ 2 x ↓ Bring down the 9. −x + 9

− − − x divided by x equals − 1. x 2 −1( x + 2) =−− x 2 . +11

(−+x9) −−−( x 2) = 11

The remainder is 11 ())x3+3 x 2 + x + 9 divided by ( x + 2 equals (x2 − x +1 ) , remainder 11.

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If the polynomial has an “ xn” term missing, add the term with a coefficient of zero.

Example (2x3 − 31 x +) ÷( x − 1 )

Rewrite (2x3 − 3 x + 1 ) as 2x3+ 0 x 2 − 31 x +

Use the method from the previous example to do the duvision.

3 2 2x÷ x = 2 x 2x2 + 2 x − 1 2xx2( − 12) = x 3 − 2 x 2 x−12 x3 + 0 x 2 −+ 3 x 1 2x3− 0 x 2−(2x3 − 2 x 2 ) = 2 x 2 2x3− 2 x 2 ↓ ↓

2 2x2 ÷ x = 2 x 2x− 3 x ↓ 2xx( − 12) = x2 − 2 x 2x2 − 2 x ↓ 2x2 − 3 x−(2x2 − 2 x ) = − x −x + 1 −x + 1 −x ÷ x =− 1 0 −1( x − 1) =−+ x 1 −+x1− ( x += 1) 0 Remainder is 0

(2x3 − 31 x +) ÷( x − 1 ) = 2x2 + 2 x − 1 , exactly. (Remainder = 0)

Exercise 1 Divide the first polynomial by the second.

(a) 2x3− 6 x 2 + 52 x + , x − 2 (b) 3x3+ 13 x 2 + 6 x − 12 , 3x + 4

(c) 3x3 + x − 1 , x +1 (d) x3 +2 x − 3 , x −1

(e) 2x4+ 5 x 3 + x + 3 , 2x + 1 (f) 2x3+ x 2 + 512 x + , 2x + 3

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Factor If P(x) is a polynomial in x and theorem P(a) = 0 then ( x −−− a) is a factor of P(x)

Solving cubic We can use the factor theorem to find one factor of a cubic function, and then equations use polynomial long division to find the remaining factor(s). (Sometimes it is possible to find all solutions by finding three values of x for which P(x) = 0 ). A cubic equation has a maximum of three distinct solutions.

Example Solve the equation x3− 5x 2 − 2x + 24 = 0

Look for a value of x that makes P( x) = 0.

Try x = 1 first, then work up through the factors of 24.

Try x = 1 P(1) =18 x − 1 is not a factor

Try x = 2 P(2) = 8 x − 2 is not a factor

Try x = − 2 P( −2) =0 therefore P( −2) = 0 x + 2 is a factor

Next divide x3− 5x 2 − 2x + 24 by ( x+ 2), using the method of polynomial long division shown above. The remainder should be zero.

x2 − 7 x + 12

3 2 x+2 x − 5 x −+ 2 x 24 x3+ 2 x 2

− 7x2 − 2 x

− 7x2 − 14 x 12x + 24

+ 12x 24 0

The factors of x3− 5x 2 − 2x + 24 are ( x + 2) and ( x2 − 7x + 12) The factors of (x2 − 7x + 12) are ( x − 3) and ( x − 4).

Therefore x3− 5x 2 − 2x + 24 has factors ( x+ 2), (x − 3) and ( x − 4) and

x3− 5x 2 − 2x + 24 = 0 has solutions x = − 2, x = 3 and x = 4.

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Exercise 2 Solve the following equations. (a) x3+ 7x 2 +11 x + 5 = 0 (b) 4x3+ 2x 2 − 2x = 0

(c) −x3− 3x 2 + x + 3 = 0 (d) x3 − 7x − 6 = 0

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Answers (a) (b) (c) Exercise 1 2x2 − 2 x + 1, remainder 4 x2 +3 x − 2, remainder − 4 3x2 − 3 x + 4, remainder − 5

(d) (e) (f) x2 + x + 3, remainder 0 x3+2 x 2 − x + 1, remainder 2 x2 − x + 4, remainder 0

Exercise 2 (a) x = −1, x = −5 (b) x = 0, x = −1, x = 1 (c) x = 1, x = −1, x = −3 2

(d) x = −1, x = −2, x = 3

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