Video Transcript
Factorize fully 12π₯ squared π¦ squared minus 26π₯ squared π¦ cubed plus 12π₯ squared π¦ raised to the fourth power.
In this question, we are asked to fully factor an algebraic expression in two variables with three terms. The first thing we want to do is check for any common factors among the terms, since we can factor these out of the expression immediately.
Letβs start with the coefficients. The highest common factor of 12 and 26 is two. So the highest common factor of the coefficients is two. We can follow the same process for the variables. We see that all three terms share a factor of π₯ squared. So we can take this factor out of the expression. We can also see that all three terms share a factor of π¦ squared. So we can also take this factor out. We call two π₯ squared π¦ squared the highest common factor of the three terms. This gives us two π₯ squared π¦ squared multiplied by six minus 13π¦ plus six π¦ squared. We usually write expressions with the powers of the variables getting smaller. So we will rewrite the factor as shown.
We are not done yet. We need to check if we can factor the quadratic into linear factors. To do this, we note that the quadratic is nonmonic. So we want to find two numbers whose product is six times six, which is 36, and whose sum is negative 13. By considering the factor pairs of 36, we can see that negative four times negative nine is 36 and negative four plus negative nine is negative 13. This allows us to apply factoring by grouping by using these two values to rewrite the π¦-term. We want to factor two π₯ squared π¦ squared multiplied by six π¦ squared minus four π¦ minus nine π¦ plus six.
We now factor each pair of terms separately. We can take a factor of two π¦ out of the first two terms to get two π¦ times three π¦ minus two. We then need to factor the last pair of terms to have another factor of three π¦ minus two. We can see that this occurs if we take out a factor of negative three from the terms. Remember, we need to multiply this entire expression by two π₯ squared π¦ squared. We now want to take out the shared factor of three π¦ minus two. This gives us the following expression. There are no shared factors in the terms of the linear factors, so we cannot factor any further. Hence, our answer is two π₯ squared π¦ squared times two π¦ minus three times three π¦ minus two.