### Video Transcript

π of π₯ equals four π₯ squared minus 10π₯ plus seven. Express π of π₯ over two plus one in the form ππ₯ squared plus ππ₯ plus π.

So first of all, to solve this problem, we need to see well, what does it actually
mean? So we got π 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 π of
π₯.

So weβre actually gonna substitute that in where weβd have π₯ in our original
function. So therefore, if we actually do that, what we get is that π 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 weβre gonna need to do is actually first of all expand our bracket where
weβve got π₯ over two plus one all squared. So our first term is actually gonna be π₯ squared over four. And thatβs because weβve 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 thatβs because we had π₯
over two multiplied by one. So now, our next term is also π₯ over two and itβs because we get positive one
multiplied by π₯ over two because itβs 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 weβve got π₯ squared over four plus π₯ over two plus π₯ over two plus one.

So letβs put this back into our expression. So then, we have π 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
itβs 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
thatβs because we had negative 10 multiplied by positive one. And then, finally, we have our plus seven on the end.

Okay, great, now, letβs 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 thatβs cause we multiplied our four by π₯ squared over four. Then, we have plus eight π₯ over two and thatβs 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 thatβs 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, weβre left with π 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 thatβs because we have
four π₯ squared divided by four. Well, four divided by four is just one. So weβre left with π₯ squared. And then, this is plus four π₯ and thatβs because we had eight π₯ over two. Well, eight divided by two is four. So weβre 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 π of π₯ is equal to four π₯ squared minus 10π₯ plus
seven, then π of π₯ over two plus one is gonna be equal to π₯ squared minus π₯ plus
one in the form ππ₯ squared plus ππ₯ plus π. 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 π of π₯ over two plus one is equal to
π₯ squared minus π₯ plus one.