# Question Video: Stating the Parity of a Polynomial Function Mathematics • 12th Grade

Is the function π(π₯) = 8π₯Β³ + 8π₯, β π₯ β β even, odd, or neither even nor odd?

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### Video Transcript

Is the function π of π₯ equals eight π₯ cubed plus eight π₯, for all real values of π₯, even, odd, or neither even nor odd?

We recall that any real-valued function is even if π of negative π₯ is equal to π of π₯. A function is odd, on the other hand, if π of negative π₯ is equal to the negative of π of π₯. In order to work out whether this function, eight π₯ cubed plus eight π₯, is even or odd, we need to find an expression for π of negative π₯. We do this by replacing all our values of π₯ in the initial function with negative π₯.

Cubing negative π₯ means multiplying negative π₯ by negative π₯ by negative π₯. This is equal to negative π₯ cubed. This means that eight multiplied by negative π₯ cubed is equal to negative eight π₯ cubed. Multiplying eight by negative π₯ gives us negative eight π₯, as multiplying a positive by a negative gives a negative term. As both these terms are negative, we can factor or factorize out negative one. This can be rewritten as negative eight π₯ cubed plus eight π₯.

As the part inside the parentheses or bracket, eight π₯ cubed plus eight π₯, is equal to our initial function π of π₯, then π of negative π₯ is equal to negative π of π₯. We can therefore conclude that the function eight π₯ cubed plus eight π₯, for all real values of π₯, is odd.