Solve:
•x^4 - 58x^2 + 441 = 0
^Quadratic equation^
Answers
The first term is, x4 its coefficient is 1 .
The middle term is, -58x2 its coefficient is -58 .
The last term, "the constant", is +441
Step-1 : Multiply the coefficient of the first term by the constant 1 • 441 = 441
Step-2 : Find two factors of 441 whose sum equals the coefficient of the middle term, which is -58 .
-441 + -1 = -442
-147 + -3 = -150
-63 + -7 = -70
-49 + -9 = -58 That's it
Step-3 : Rewrite the polynomial splitting the middle term using the two factors found in step 2 above, -49 and -9
x4 - 49x2 - 9x2 - 441
Step-4 : Add up the first 2 terms, pulling out like factors :
x2 • (x2-49)
Add up the last 2 terms, pulling out common factors :
9 • (x2-49)
Step-5 : Add up the four terms of step 4 :
(x2-9) • (x2-49)
Which is the desired factorization
x⁴ - 58x² + 441= 0
let x² be y
=> y² - 58y + 441 = 0
y² - 49y - 9y + 441 = 0
y(y - 49) - 9(y - 49) = 0
(y - 49) (y - 9) = 0
=> y = 49,9
x² = 49
=> x = ± 7;
x² = 9
=> x = ± 3;
Hence found
Hope it helps you!
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