Everything Maths Grade 12

CHAPTER 5. FACTORISING CUBIC POLYNOMIALS 5.4

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5.4 Solving Cubic Equations EMCAS

Once you know how tofactorise cubic polynomials, it is also easy to solve cubic equations of the kind

ax^3 +bx^2 +cx +d = 0

Example 5: Solution of Cubic Equations

QUESTION

Solve

6 x^3 − 5 x^2 − 17 x + 6 = 0.

SOLUTION

Step 1 : Find one factor using the Factor Theorem

Try

f (1) = 6(1)^3 − 5(1)^2 − 17(1) + 6 = 6− 5 − 17 + 6 =− 10

Therefore (x− 1) is NOT a factor.

Try

f (2) = 6(2)^3 − 5(2)^2 − 17(2) + 6 = 48− 20 − 34 + 6 = 0

Therefore (x− 2) IS a factor.

Step 2 : Division by inspection

6 x^3 − 5 x^2 − 17 x + 6 = (x− 2)( )

The first term in the second bracket must be 6 x^2 to give 6 x^3 if one works back-

wards.

The last term in the second bracket must be− 3 because− 2 ×−3 = +6.

So we have 6 x^3 − 5 x^2 − 17 x + 6 = (x− 2)(6x^2 +?x− 3).

Now, we must find the coefficient of the middleterm (x).

(−2)(6x^2 ) gives− 12 x^2. So, the coefficient of the x-term must be 7.

So, 6 x^3 − 5 x^2 − 17 x + 6 = (x− 2)(6x^2 + 7x− 3).