Question

Solve. a) \((x+1)^{2}(x+2)=0\) b) \(x^{3}-1=0\) c) \((x+4)^{3}(x+2)^{2}=0\)

Short answer

a) \x = -1, -2\; b) \x = 1\; c) \x = -4, -2\.

Step-by-step solution

Recognize the Zero-Product Property

To solve equations like those given, use the Zero-Product Property. This property states that if a product of factors is zero, at least one of the factors must be zero.

Solve Part (a)

Given \((x+1)^{2}(x+2)=0\): Set each factor to zero and solve for \x: (i) \(x + 1 = 0\) leads to \(x = -1\) (ii) \(x + 2 = 0\) leads to \(x = -2\). So, the solutions for (a) are \x = -1\text{ and }x = -2\.

Solve Part (b)

Given \(x^{3}-1=0\):Rewrite as \(x^{3} = 1\). Take the cube root of both sides: \(x = \root{3}\big(1\big) = 1\). So, the solution for part (b) is \x = 1\.

Solve Part (c)

Given \((x+4)^{3}(x+2)^{2}=0\):Set each factor to zero and solve for \x: (i) \(x + 4 = 0\), which gives \(x = -4\). (ii) \(x + 2 = 0\), which gives \(x = -2\). So, the solutions for part (c) are \x = -4 \text{ and } x = -2\.

Key concepts

Solving Polynomial Equations

Solving polynomial equations often involves finding the values of the variable that make the equation true. A polynomial equation is a mathematical expression involving a sum of powers in one or more variables multiplied by coefficients. For example, the equations we are solving in this exercise are polynomial equations.

To solve these types of equations, one commonly used method is factoring, which breaks down the polynomial into simpler components. These components can be set equal to zero to find the solutions.

Factoring

Factoring is a crucial step in solving polynomial equations. It involves breaking down a polynomial into a product of smaller polynomials or numbers that when multiplied together give back the original polynomial. In our exercise:

• For part (a), \((x+1)^{2}(x+2)=0\) is already factored.
• For part (b), \({x^3}-1=0\) can be rewritten as \({(x-1)(x^2+x+1)=0\). Then, you solve each factor.
• For part (c), \({(x+4)^{3}(x+2)^{2}=0\) is also already factored.

This helps significantly simplify the equation and makes it easier to solve.

Cube Roots

In cases like part (b) where we have a cube, we use cube roots. The cube root of a number \({x\) is a number which, when multiplied by itself three times, equals x. For example, finding the cube root of 27 means finding the number that when multiplied by itself three times equals 27. The cube root is usually denoted as \({\root{3}{x}\).

In our example, \({x^3 = 1\) becomes \({x = \root{3}{1} = 1\). Hence, the solution for part (b) is \({x = 1\).

Quadratic Equations

Quadratic equations are a specific type of polynomial equation where the highest power of the variable is two. These equations take the standard form of \({ax^2 + bx + c = 0\), where \({a\), \({b\), and \({c\) are constants.

In our example, although we didn't directly solve a typical quadratic equation, we used similar concepts. Particularly, when we set each factor to zero, some of them resulted in quadratic forms. Solving each factor individually simplifies even quadratic equations.

For example, in part (c), \((x+4)^{3}(x+2)^{2}=0\), the steps to solve each factor are the same process involved in solving quadratic equations. This shows the versatility of these methods in different types of polynomial equations.