Video: Using Synthetic Substitution to Evaluate Polynomial Functions

Consider the function 𝑓(π‘₯) = π‘₯⁴ βˆ’ 9π‘₯Β³ + 3π‘₯Β² βˆ’ 7π‘₯ + 12. Use synthetic division to find the quotient 𝑄(π‘₯) and the remainder 𝑅 satisfying 𝑓(π‘₯) = 𝑄(π‘₯)(π‘₯+2) + 𝑅. Find 𝑓(βˆ’2).

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Video Transcript

Consider the function 𝑓 of π‘₯ equals π‘₯ to the fourth power minus nine π‘₯ cubed plus three π‘₯ squared minus seven π‘₯ plus 12. Use synthetic division to find the quotient 𝑄 of π‘₯ and the remainder 𝑅, satisfying 𝑓 of π‘₯ equals 𝑄 of π‘₯ times π‘₯ plus two plus 𝑅. Then we should find 𝑓 of negative two.

So we will be using synthetic division to find the quotient 𝑄 of π‘₯ and the remainder 𝑅. So when we divide, we’ll have this quotient. And if there’s anything left over, we will have the remainder. So notice we want to satisfy the equation 𝑓 of π‘₯ equals 𝑄 of π‘₯, our quotient, times π‘₯ plus two plus 𝑅, the remainder. So we must divide by the π‘₯ plus two. So we need to take our function and divide by π‘₯ plus two.

Now to do this with synthetic division, we need to take π‘₯ plus two and set it equal to zero and then divide by that number. So we need to subtract two from both sides of the equation. And we find that π‘₯ is equal to negative two. So this will go on the outside. And then we draw, looks like a long L. And at the top on the inside, we put the coefficients of our function. So we have one, negative nine, three, negative seven, and 12. And they need to go in this order, descending order with the exponents.

And it’s already in that. And if we had anything missing, say there wasn’t an π‘₯ squared, we would need to put a zero in place of it. So- so, we write these numbers inside here. So our first step is to bring down the one. And then we take one times negative two. One times negative two is negative two, and we write that here. And now we add negative nine and negative two together, which is negative 11. And now we repeat the process.

Negative 11 times negative two is 22. And 22 plus three is 25. 25 times negative two is negative 50. Negative 50 plus negative seven is negative 57. And now negative 57 times negative two is positive 114. And 114 plus 12 is 126. So these numbers are in this order for a reason. All the way to the right is the remainder. And then next to that to the left will be the constant. And then next to that will be the coefficient of π‘₯ and then the coefficient of π‘₯ squared and then the coefficient of π‘₯ cubed.

Therefore, 𝑄 of π‘₯ is equal to π‘₯ cubed minus 11π‘₯ squared plus 25π‘₯ minus 57, and the remainder is equal to 126. So lastly, we need to find 𝑓 of negative two, which means we need to plug negative two into our function. So everywhere that there is an π‘₯, we replace it with the negative two and now we evaluate.

Negative two to the fourth power is 16. And then negative two cubed is negative eight. So we don’t wanna multiply negative nine and negative two together. We must first take care of all of the exponents before we ever multiply. So negative two cubed is negative eight. And then next, we had negative two squared, which is positive four. Now the next one is negative two technically to the first power. So it’s just negative two.

So we already took negative seven times negative two to be positive 14. Now we can multiply these together. Negative nine times negative eight is positive 72. Three times four is 12. So we have 16 plus 72 plus 12 plus 14 plus 12, which is equal to 126. And yes, there is a relationship between these, but that will be for another lesson.

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