In Exercises 39–44, factor by introducing an appropriate substitu...

Question

In Exercises 39–44, factor by introducing an appropriate substitution.

x⁴ − 4x² − 5

Accepted Answer

Hello, everyone. We are gonna perform factorization using substitution on the expression X raised to the fourth power plus three X squared minus 10. Our answer choices are a open parentheses. X squared minus two closed parentheses multiplied by open parentheses, X squared minus five closed parentheses, B open parentheses, X squared minus two closed parentheses, multiplied by open parentheses. X squared plus five closed parentheses. C open parentheses, X squared plus two closed parentheses multiplied by open parentheses. X squared minus five closed parentheses. D open parentheses, X squared plus two closed parentheses, multiplied by open parentheses. X squared plus five closed parentheses, rewriting the original expression x-rays to the 4th power plus three X squared minus 10. The reason our problem is asking us to use substitution is because at this point, we cannot factor this easily into our solutions because we have an X race, the fourth power and an X squared. So we're gonna substitute a different variable in for X squared to get this looking a little bit more like our traditional trinomial. I'm gonna use the variable of U and U is gonna represent X squared. So I'm gonna rewrite this expression with U in for X squared. So X raised the fourth power is the same thing as saying X squared squared. So this will now be U squared plus 3, x squared is now gonna be three U minus 10. This looks a lot more like the traditional trinomial we're used to factoring. So now I'm gonna factor this trinomial into two sets of parentheses. We know our factorization results in two binomials. So the first term in each binomial is gonna be U because that will multiply back together to U squared. And then I need two values that not only multiply to negative but add to a positive three. Since they multiply to a negative, I know I have opposite signs. So I have a positive and a negative. So one of my parentheses has a plus sign one has a minus sign if it helps to make a list off to the side of things that will multiply to negative 10. So you can check to see if they add to a positive three, go right ahead. So it could be negative one, positive 10, positive, one, negative 10, -2, positive 5, positive 2, -5. Those look like all my options. Negative one plus positive 10 does not add to a positive three, one plus a negative 10 does not add to a positive three, negative two plus five does add to a positive three. So I'm gonna fill that in here. So the factorization at this step is open parentheses, U plus five closed parentheses multiplied by open parentheses, U minus two closed parentheses. Now that I have this as factored as I can, I'm gonna replace you with the X squared that we used before. So I'm going to undo my substitution. So in the first set of parentheses I have now X squared plus five closed parentheses. Next set of parentheses would be X squared minus two closed parentheses. This is my solution. This is my factorization. There's nothing else I can combine nothing else I can factor. And when I look at my answer choices, recall that multiplication can be done in either order. So our factors have to be there, but they don't need to be in this precise order. Because answer choice B is our solution and that's open parentheses. X squared minus two closed parentheses, multiplied by open parentheses. X squared plus five closed parentheses. Same values. Different order have a nice day.

In Exercises 39–44, factor by introducing an appropriate substitution. x⁴ − 4x² − 5

Concetti chiave

Factoring Polynomials

Substitution Method

Quadratic Equations

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