Algebra II : Intermediate Single-Variable Algebra
Study concepts, example questions & explanations for Algebra II
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Example Questions
Example Question #9 : Write A Polynomial Function From Its Zeros
Which of the following equations belongs to a quadratic with zeros at x = 2 and x = 7?
The answer is 
A quadratic with zeros at x = 2 and x = 7 would factor to 
However, if you multiply 
Example Question #721 : Intermediate Single Variable Algebra
Which of the following equations represents a quadratic equation with zeros at x = 3 and x = -6, and that passes through point (2, -24)?
When finding the equation of a quadratic from its zeros, the natural first step is to recreate the factored form of a simple quadratic with those zeros. Putting aside the coefficient, a quadratic with zeros at 3 and -6 would factor to:
So you know that a possible quadratic for these zeros would be:
Now you need to determine the coefficient of the quadratic, and that’s where the point (2, -24) comes in. That means that if you plug in x = 2, the result of the quadratic will be -24. So you can set up the equation:
So a = 3. When you then distribute that coefficient of 3 across the original simple quadratic, you have:
So your quadratic is
Example Question #11 : Write A Polynomial Function From Its Zeros
Which of the following equations belongs to a quadratic with zeros at 
A quadratic with zeros at 2 and 7 would factor to 
However, if you multiply 
Example Question #13 : Write A Polynomial Function From Its Zeros
Which of the following represents a quadratic with zeros at 5 and 2 and that passes through point (1, 16)?
When you're finding the equation of a quadratic given its zeros, a good start is to ignore the coefficient (you'll return to it later) and to construct the factored form of the quadratic using the zeros. Here with zeros of 5 and 2, that factored form would be:
That then means you can expand it to:
From here, you can see that the answer choices all maintain the same general relationship as this simple quadratic, but have different coefficients. This is where you can use the point (1, 16) to your advantage. This means that if you plug in 
Where 
Meaning that:
So you know that 
Then you can multiply that coefficient across your simple quadratic to get the answer:
Becomes:
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