Video: Pack 5 β€’ Paper 2 β€’ Question 13

Pack 5 β€’ Paper 2 β€’ Question 13

04:08

Video Transcript

𝑓 of π‘₯ equals four π‘₯ squared minus 10π‘₯ plus seven. Express 𝑓 of π‘₯ over two plus one in the form π‘Žπ‘₯ squared plus 𝑏π‘₯ plus 𝑐.

So first of all, to solve this problem, we need to see well, what does it actually mean? So we got 𝑓 of π‘₯ over two plus one. So how do we actually do that? What we do to our original function? Well, what we actually do is we substitute in π‘₯ over two plus one for π‘₯ in 𝑓 of π‘₯.

So we’re actually gonna substitute that in where we’d have π‘₯ in our original function. So therefore, if we actually do that, what we get is that 𝑓 of π‘₯ over two plus one is gonna be equal to four multiplied by π‘₯ over two plus one all squared because that was where our π‘₯ was minus 10 multiplied by π‘₯ over two plus one plus seven.

So now, what we’re gonna need to do is actually first of all expand our bracket where we’ve got π‘₯ over two plus one all squared. So our first term is actually gonna be π‘₯ squared over four. And that’s because we’ve got π‘₯ over two multiplied by π‘₯ over two. If you multiply fractions, you multiply the numerators. So π‘₯ multiplied by π‘₯ is π‘₯ squared. And you multiply the denominators. So two multiplied by two is four.

Then, our next term is gonna be positive π‘₯ over two and that’s because we had π‘₯ over two multiplied by one. So now, our next term is also π‘₯ over two and it’s because we get positive one multiplied by π‘₯ over two because it’s the second term in the first bracket multiplied by the first term in the second bracket. And then, finally, we have one multiplied by one, which gives us positive one. So we’ve got π‘₯ squared over four plus π‘₯ over two plus π‘₯ over two plus one.

So let’s put this back into our expression. So then, we have 𝑓 of π‘₯ over two plus one is equal to four multiplied by π‘₯ squared over four plus π‘₯ over two plus π‘₯ over two plus one. And then if we expand the second bracket, we get the first term as negative 10π‘₯ over two because we had negative 10 multiplied by π‘₯ over two. Be careful here with negative signs when you are multiplying out by bracket because it’s the sign before the term in front of the bracket. And then, we have minus 10 as our second term when we expand the second bracket and that’s because we had negative 10 multiplied by positive one. And then, finally, we have our plus seven on the end.

Okay, great, now, let’s see if we can actually simplify this. So first of all, we can say that this is gonna be equal to four π‘₯ squared over four. And that’s cause we multiplied our four by π‘₯ squared over four. Then, we have plus eight π‘₯ over two and that’s because well, we had π‘₯ over two plus π‘₯ over two which gives us two π‘₯ over two. Then, multiplying that by four gives us eight π‘₯ over two and then plus four because we had four multiplied by one which gives us four. We then have minus five π‘₯ and that’s because we had negative 10π‘₯ over two or 10 divided by two is five and, then, finally, minus three because we had negative 10 plus seven, which gives us negative three.

Okay, great, so now, we’re left with 𝑓 of π‘₯ over two plus one equals four π‘₯ squared over four plus eight π‘₯ over two plus four minus five π‘₯ minus three. Okay, so we can actually now simplify again. So our first term is actually just gonna be π‘₯ squared and that’s because we have four π‘₯ squared divided by four. Well, four divided by four is just one. So we’re left with π‘₯ squared. And then, this is plus four π‘₯ and that’s because we had eight π‘₯ over two. Well, eight divided by two is four. So we’re left with positive four π‘₯ and then plus four minus five π‘₯ minus three.

Okay, onto the final stage because we can do one last part. We can actually combine like terms. So therefore, we can say that if 𝑓 of π‘₯ is equal to four π‘₯ squared minus 10π‘₯ plus seven, then 𝑓 of π‘₯ over two plus one is gonna be equal to π‘₯ squared minus π‘₯ plus one in the form π‘Žπ‘₯ squared plus 𝑏π‘₯ plus 𝑐. And we got this final answer because positive four π‘₯ minus five π‘₯ gives us negative π‘₯ and positive four minus three give us positive one.

So therefore, as we said, the final answer is 𝑓 of π‘₯ over two plus one is equal to π‘₯ squared minus π‘₯ plus one.

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