Question

Find the products-

i) (x²+1/2)(x²+1/3)(x-1/6)
ii) (p-6)(p-5)(p+10)

Using the formula-
(x+a) (x+b) (x+c) = {x³+(a+b+c)x²+ (ab+bc+ca)x +abc​

Answer 1

Step-by-step explanation:

How can we solve equations such as the cubic equation shown here?

x3 − x2 – 4x + 4 = 0

There is an extremely complicated formula for solving cubic equations. Some calculators have this formula built in and therefore can be used to solve cubic equations.

We are going to learn how these equations can be solved by factorising. If the equation has solutions that are integers a, b and c then we can factorise the equation as follows:

x3 − x2 – 4x + 4 = (x − a)(x − b)(x − c) = 0

Multiplying the brackets together we see that the constant term, 4, must be the number we get when we multiply a, b and c together.

abc = 4

All the solutions a, b and c must be factors of 4 so there are not many whole numbers that we need to consider.

We have only the following possibilities:

±1, ±2 and ±4

We’ll examine each of these numbers to find which ones are solutions of the equation.

f(1) = 13 − 12 – 4×1 + 4 = 0 1 is a solution

f(−1) = (−1)3 − (−1)2 – 4×(−1) + 4 = 6

f(2) = 23 − 22 – 4×2 + 4 = 0 2 is a solution

f(−2) = (−2)3 − (−2)2 – 4×(−2) + 4 = 0 −2 is a solution

We have now found three solutions so we don’t need to try 4 and −4 as a cubic equation has a maximum of three solutions.

These three numbers give us the values of a, b and c and we can factorise the equation.

x3 − x2 – 4x + 4 = (x − 1)(x − 2)(x + 2) = 0

This method involves finding integers that are factors of ( can be divided into ) the constant term and then testing whether these integers are solutions of the equation.

Unfortunately we cannot assume that the solutions of a third degree equation are all integers.

However, if we can find one integer solution, lets say it is x = a then, by the remainder theorem , we know that (x − a) is a factor of the equation. We can find another factor, a quadratic factor, by division. We can then solve the quadratic equation by using the formula for solving quadratics.

Answer 2

Step-by-step explanation:

i)

Using the formula-

(x+a) (x+b) (x+c) = {x³+(a+b+c)x²+ (ab+bc+ca)x +aabc

(x²+1/2)(x²+1/3)(x-1/6)= { (x^2)^3+(1/2+1/3-1/6)x^2+(1/2×1/3 + 1/3×(-1/6)+(-1/6×1/2)x+1/2×1/3×(-1/6)}

= {x^6+((5-1)/6 )x^2 +(1/6-1/18-1/12)x-1/36}

= { x^6 + 2/3x^2+(6-2-3)/36x - 1/36}

= { x^6 + 2/3x^2+ 1/36x - 1/36}