Video: AQA GCSE Mathematics Higher Tier Pack 2 β€’ Paper 3 β€’ Question 18

Expand and simplify (2π‘₯ βˆ’ 4)(5π‘₯Β² + 3) βˆ’ 3(βˆ’2π‘₯Β³ + π‘₯).

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Video Transcript

Expand and simplify two π‘₯ minus four times five π‘₯ squared plus three minus three times negative two π‘₯ cubed plus π‘₯.

Our first step will be to multiply out all of the brackets. Beginning with the first term, we have two π‘₯ times five π‘₯ squared, two π‘₯ times three. Notice that we’re going to be adding all of these terms that we’re multiplying together. When we move on to negative four times five π‘₯ squared, we’re adding negative four times five π‘₯ squared.

It would still be true to say subtracting four times five π‘₯ squared. However, keeping that negative with its term will prevent us from making sign mistakes later in the problem. We’ll multiply negative four times three. And then we can multiply out negative three times negative two π‘₯ cubed plus π‘₯. I’ll use addition and then say negative three times negative two π‘₯ cubed plus negative three times π‘₯.

Let’s do some multiplication here. Two π‘₯ times five π‘₯ squared equals 10π‘₯ cubed, plus two π‘₯ times three, which equals six π‘₯. Negative four times five π‘₯ squared equals negative 20π‘₯ squared. Negative four times three equals negative 12. Negative three times negative two π‘₯ cubed equals six π‘₯ cubed. Negative times a negative is a positive. And then negative three times π‘₯ equals negative three π‘₯.

In our next step, in order to simplify, we need to combine like terms. Like terms are terms whose variables and their exponents are the same. 10π‘₯ cubed and six π‘₯ cubed are like terms. We can add them together. We add them together by adding their coefficients, 10 plus six. And we get 16π‘₯ cubed. And we’re finished with those two terms.

Now that we’re finished with our π‘₯ cubed term, we’ll look for π‘₯ squared terms. There’s only one negative 20π‘₯ squared. So now we’ll say minus 20π‘₯ squared. After the π‘₯ squared term, we’ll look for any π‘₯ terms, π‘₯ to the first power. And we have six π‘₯ plus negative three π‘₯. When we add those together, we get positive three π‘₯.

Our last term will be any constants. And we only have one, negative 12. So we just bring that down. The expanded and simplified form of this expression is 16π‘₯ cubed minus 20π‘₯ squared plus three π‘₯ minus 12.

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