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in this lesson were going to talk about solving polynomial equations that are in quadratic form so heres an example x to the fourth minus 13 x squared plus 36 is equal to 0. lets find the value of x notice that the exponent of the first term is twice the value of the middle term when you see that its in quadratic form so were going to factor by substitution so lets pick a variable lets say a lets set a equal to the middle variable which is x squared so if a is x squared then a squared is x to the fourth so this is equivalent to a squared minus 13 a plus 36 which we can factor two numbers that multiply to 36 but add to negative 13. are nine negative nine that is and uh negative four so therefore we can see that a is equal to nine and a is equal to four but before i do that i want to factor it completely so before solving for a im going to replace a with x squared so therefore we have x squared minus 9 and x squared minus 4 which we could factor each of those expressions using