Question

algebra 2 honors
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midterm review #2 (unit 3 and 4)
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state the possible rational zeros for each function. then find all rational zeros.

1. ( y = x(3x - 4)(x + 1) )
2. ( y = x(2x + 3)(x - 1) )

Explanation:

Step1: Recall the Zero Product Property

The Zero Product Property states that if \( ab = 0 \), then either \( a = 0 \) or \( b = 0 \) (or both). For a function in factored form \( y=(x - a)(x - b)(x - c)\), the zeros are found by setting each factor equal to zero and solving for \( x \). Also, to find possible rational zeros, we can analyze the factors. For a function like \( y = x(ax + b)(x + c)\), the possible rational zeros come from the roots of each linear factor.

Step2: Solve for \( y = x(3x - 4)(x + 1) \)

Step 2.1: Find possible rational zeros

The factors are \( x \), \( 3x - 4 \), and \( x + 1 \).

• For \( x = 0 \), solving \( x = 0 \) gives \( x = 0 \).
• For \( 3x - 4 = 0 \), we solve \( 3x=4\) so \( x=\frac{4}{3}\).
• For \( x + 1 = 0 \), we solve \( x=- 1\).

So the possible rational zeros are \( 0,\frac{4}{3}, - 1\), and the rational zeros (since we found them by setting factors to zero) are \( x = 0\), \( x=\frac{4}{3}\), \( x=-1\).

Step3: Solve for \( y = x(2x + 3)(x - 1) \)

Step 3.1: Find possible rational zeros

The factors are \( x \), \( 2x + 3 \), and \( x - 1 \).

• For \( x = 0 \), solving \( x = 0 \) gives \( x = 0 \).
• For \( 2x+3 = 0 \), we solve \( 2x=-3\) so \( x =-\frac{3}{2}\).
• For \( x - 1 = 0 \), we solve \( x = 1\).

So the possible rational zeros are \( 0,-\frac{3}{2},1\), and the rational zeros are \( x = 0\), \( x=-\frac{3}{2}\), \( x = 1\).

Answer:

1. For \( y=x(3x - 4)(x + 1) \):

• Possible rational zeros: \( 0,\frac{4}{3},-1 \)
• Rational zeros: \( x = 0\), \( x=\frac{4}{3}\), \( x=-1 \)

1. For \( y=x(2x + 3)(x - 1) \):

• Possible rational zeros: \( 0,-\frac{3}{2},1 \)
• Rational zeros: \( x = 0\), \( x=-\frac{3}{2}\), \( x = 1\)