Question

factor and find all real roots:
$x^3 + 9x^2 + 20x = 0$

Explanation:

Step1: Factor out the common term

First, we can factor out the greatest common factor (GCF) from the left - hand side of the equation \(x^{3}+9x^{2}+20x = 0\). The GCF of \(x^{3}\), \(9x^{2}\), and \(20x\) is \(x\). So we have:
\(x(x^{2}+9x + 20)=0\)

Step2: Factor the quadratic expression

Next, we factor the quadratic expression \(x^{2}+9x + 20\). We need to find two numbers that multiply to \(20\) and add up to \(9\). The numbers \(4\) and \(5\) satisfy this condition because \(4\times5 = 20\) and \(4 + 5=9\). So we can factor \(x^{2}+9x + 20\) as \((x + 4)(x+5)\). Then the factored form of the original equation is:
\(x(x + 4)(x + 5)=0\)

Step3: Find the real roots

To find the real roots, we set each factor equal to zero and solve for \(x\):

• For the factor \(x\): \(x=0\)
• For the factor \(x + 4\): \(x+4=0\), which gives \(x=-4\)
• For the factor \(x + 5\): \(x + 5=0\), which gives \(x=-5\)

Answer:

The real roots of the equation \(x^{3}+9x^{2}+20x = 0\) are \(x = 0\), \(x=-4\), and \(x=-5\)