Question Video: Evaluating a Polynomial Function for Variables and Algebraic Expressions Mathematics

Consider the polynomial function 𝑓(π‘₯) = 2π‘₯Β³ + 5π‘₯Β² βˆ’ 7π‘₯ + 10. Evaluate 𝑓(π‘₯Β² + 1).

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Video Transcript

Consider the polynomial function 𝑓 of π‘₯ equals two π‘₯ cubed plus five π‘₯ squared minus seven π‘₯ plus 10. Evaluate 𝑓 of π‘₯ squared plus one.

So in order to evaluate, instead of having π‘₯ being plugged into our function, we need to plug in π‘₯ squared plus one. So everywhere that there is an π‘₯ in our function, we need to replace it with π‘₯ squared plus one β€” here and also here. So we need to plug in π‘₯ squared plus one into three different places and then simplify.

So here we’ve taken our function and instead of π‘₯, we’ve replaced it with π‘₯ squared plus one. Let’s begin by rewriting this first piece. Instead of writing π‘₯ squared plus one to the third power, that’s the same thing as π‘₯ squared plus one times π‘₯ squared plus one times π‘₯ squared plus one. This next piece is very similar. Instead of the square, we can rewrite it this way. For this third piece, we actually need to distribute. Negative seven times π‘₯ squared is negative seven π‘₯ squared and negative seven times one is negative seven. And then, lastly, we have this plus 10.

In our next steps, there will be lots of multiplying. We will begin by foiling these two parentheses, multiplying them together. π‘₯ squared times π‘₯ squared is π‘₯ to the fourth power. π‘₯ squared times one is equal to π‘₯ squared. One times π‘₯ squared is equal to π‘₯ squared. And one times one equals one. And π‘₯ squared plus π‘₯ squared is equal to two π‘₯ squared.

On our next step, we will multiply by this last π‘₯ squared plus one, but not yet. Next, we need to take π‘₯ squared plus one times π‘₯ squared plus one. But we actually already know what that’s equal to because we’ve just completed it. It’s equal to π‘₯ to the fourth plus two π‘₯ squared plus one. We bring down our negative seven π‘₯ squared and then negative seven plus 10 is three.

Next, we need to multiply these together: π‘₯ to the fourth times π‘₯ squared is equal to π‘₯ to the sixth because we add our exponents. π‘₯ to the fourth times one is π‘₯ to the fourth. Two π‘₯ squared times π‘₯ squared is two π‘₯ to the fourth power. Two π‘₯ squared times one is equal to two π‘₯ squared. One times π‘₯ squared is equal to π‘₯ squared. And one times one equals one. And we can simplify this because π‘₯ to the fourth plus two π‘₯ to the fourth is three π‘₯ to the fourth and two π‘₯ squared plus π‘₯ squared is equal to three π‘₯ squared.

Now, our next step would be to distribute the five. And five times π‘₯ to the fourth power is five π‘₯ to the fourth power. Five times two π‘₯ squared is equal to 10π‘₯ squared. And five times one is equal to five. And we bring down our last two terms. We can put the five and three together to make eight. And we can also put together the 10π‘₯ squared and the negative seven π‘₯ squared. That is equal to three π‘₯ squared.

So now lastly, we need to distribute the two. Two times π‘₯ to the sixth is two π‘₯ to the sixth power. Two times three π‘₯ to the fourth power is equal to six π‘₯ to the fourth power. Two times three π‘₯ squared is equal to six π‘₯ squared. And two times one is equal to two. And after bringing down our other terms, we can combine like terms now.

We can go ahead and write two π‘₯ to the sixth power because there’s no other π‘₯ to the sixth power that it can combine with. Six π‘₯ to the fourth power plus five π‘₯ to the fourth power is equal to 11π‘₯ to the fourth power. Six π‘₯ squared plus three π‘₯ squared is equal to nine π‘₯ squared. And then, two plus eight is equal to 10.

So our final answer is two π‘₯ to the sixth power plus 11π‘₯ to the fourth power plus nine π‘₯ to the second power plus 10.

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