### Video Transcript

Factor fully 24π₯ to the fifth power minus 34π₯ cubed plus 12π₯.

In order to factor this, we first want to check and see if any of these terms have common factors. We notice that all three coefficients are even, which means they have a factor of two. And then we see that they all have at least one π₯, which means we can factor out an π₯. If we factor two π₯ out of 24π₯ to the fifth power, we get 12π₯ to the fourth power. If we factor two π₯ out of negative 34π₯ cubed, we get negative 17π₯ squared. And factoring out a two π₯ from 12π₯ leaves us with positive six.

Here, weβre dealing with a higher-order polynomial, and so we probably need a new strategy to help us factorize fully. And one of those strategies is to use π’-substitution. If we let π’ equal π₯ squared, then we can rewrite 12π₯ to the fourth power as 12 times π’ squared. π’ squared is equal to π₯ squared squared, which is π₯ to the fourth power. Our second term then becomes 17π’, and the six stays the same. We do this because weβre more used to working with polynomials with a degree of two.

This doesnβt fully solve our problem though because we still have a leading coefficient of 12, which means weβll have lots of factors to consider. Weβll have to consider the factors of 12 and the factors of six. With 12, we have one and 12, two and six, or three and four. With six, we have one and six or two and three. If we start with the factors one and 12, weβll have π’ times 12π’ and then we could consider one and six. We have to multiply the outsides and the insides, and they have to add together to equal negative 17. This means that both values should be negative because they need to multiply together to equal positive six. In this case, weβd have negative six π’ minus 12π’, which is close. Itβs negative 18π’, but itβs not the exact factors.

If we tried to switch the six and one, weβd be multiplying negative six by 12π’, which is negative 72π’, far too much. We can move on to two and three. In this case, weβd have negative three π’ minus 24π’. Negative 27π’ doesnβt work. Now, we could systematically keep working through all the options until we find the correct one. Another strategy is to look at these factors and see if you can find pairs that would add up to negative 17. We know that the factors of six will all need to be negative. So we might think six times negative three is negative 18 plus two times negative two, which is negative four. But negative 18 minus four is not the right value.

But what about three times negative three, which is negative nine, and four times negative two, which is negative eight? Negative nine plus negative eight does equal negative 17. And that means we need three π’ to be multiplied by negative three, so we put those on the outside. And we need four π’ to be multiplied by negative two, so we put both of those on the inside. Itβs probably worth quickly checking our work here. Three π’ times four π’ is 12π’ squared. Three π’ times negative three is negative nine π’. Negative two times negative four π’ is negative eight π’. And negative two times negative three is positive six, which does equal 12π’ squared minus 17π’ plus six.

Now that we have this factorized form and it doesnβt appear that we can factor anything else, we want to substitute back in π₯ squared for π’, so that we have two π₯ times three π₯ squared minus two times four π₯ squared minus three.