Question

act check: circle the best answer.

1. the roots of a polynomial equation are 0, -2 and 5. which of the following is a factored form of the equation?

a. (x - 2)(x + 5) = 0
b. x(x + 2)(x - 5) = 0
c. x(x - 2)(x + 5) = 0
d. (x + 2)(x - 5) = 0

1. which of the following expressions is a factor of (x^3 - 125)?

a. (x - 5)
b. (x + 5)
c. (x^2 + 25)
d. (x^2 - 5x + 25)

Explanation:

Question 9

Step1: Recall root - factor relationship

If \( r \) is a root of a polynomial equation, then \( (x - r) \) is a factor of the polynomial. Given the roots are \( 0\), \( - 2\) and \(5\).

• For root \( x = 0\), the factor is \( x-0=x\).
• For root \( x=-2\), the factor is \(x-(-2)=x + 2\).
• For root \( x = 5\), the factor is \(x - 5\).

Step2: Form the factored equation

The polynomial in factored form is the product of its factors. So the factored form of the polynomial equation with roots \(0\), \(-2\) and \(5\) is \(x(x + 2)(x - 5)=0\).

Step1: Recall the difference of cubes formula

The difference of cubes formula is \(a^{3}-b^{3}=(a - b)(a^{2}+ab + b^{2})\). We know that \(125 = 5^{3}\), so \(x^{3}-125=x^{3}-5^{3}\).

Step2: Apply the difference of cubes formula

Comparing \(x^{3}-5^{3}\) with \(a^{3}-b^{3}\), we have \(a = x\) and \(b = 5\). Then, by the difference of cubes formula:
\(x^{3}-5^{3}=(x - 5)(x^{2}+5x + 25)\)
So the factors of \(x^{3}-125\) are \(x - 5\) and \(x^{2}+5x + 25\). Among the given options, \(x - 5\) (option A) and \(x^{2}+5x + 25\) (option D) are factors. But let's check the options again. Wait, the options are:
A. \(x - 5\)
B. \(x + 5\)
C. \(x^{2}+25\)
D. \(x^{2}-5x + 25\)

Wait, there is a typo in my initial thought. The correct expansion of \(x^{3}-5^{3}\) is \((x - 5)(x^{2}+5x + 25)\)? No, wait, no: \(a^{3}-b^{3}=(a - b)(a^{2}+ab + b^{2})\), so \(x^{3}-5^{3}=(x - 5)(x^{2}+5x + 25)\)? No, \(ab=x\times5 = 5x\), so \(a^{2}+ab + b^{2}=x^{2}+5x + 25\)? Wait, no, \(b = 5\), so \(a^{2}+ab + b^{2}=x^{2}+5x + 25\) is wrong. Wait, \(a=x\), \(b = 5\), so \(a^{2}+ab + b^{2}=x^{2}+5x + 25\)? No, \(b^{2}=25\), \(ab = 5x\), \(a^{2}=x^{2}\), so yes. But the option D is \(x^{2}-5x + 25\), which is wrong. Wait, I must have made a mistake. Wait, \(x^{3}-125=x^{3}-5^{3}\), and the correct factorization is \((x - 5)(x^{2}+5x + 25)\)? No, no, the correct formula is \(a^{3}-b^{3}=(a - b)(a^{2}+ab + b^{2})\), so if \(a=x\) and \(b = 5\), then \(a - b=x - 5\) and \(a^{2}+ab + b^{2}=x^{2}+5x + 25\). But the option A is \(x - 5\), which is a factor. Let's check by substituting \(x = 5\) into \(x^{3}-125\). When \(x = 5\), \(5^{3}-125=125 - 125=0\), so by the factor theorem, if \(f(x)=x^{3}-125\), then \(f(5) = 0\), which means \((x - 5)\) is a factor.

Answer:

B. \(x(x + 2)(x - 5)=0\)

Question 10