Determine: a) The function f(x)=x4+1 into the product of irreducible quadratics. b) The zeros of f(x)=x4+1 by finding the zeros of each irreducible quadratic.

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Determine:  a) The function f(x)=x4+1 into the product of irreducible quadratics.  b) The zeros of f(x)=x4+1 by finding the zeros of each irreducible quadratic.

jean2098 · Accepted Answer

Step 1  The given function is f(x)=x4+1  Determine the factor of the above mentioned function as follows,  f(x)=x4+1  =x4+2x2+1−2x2  =(x2)2+2×x2×1+12−2x2  =(x2+1)2−2x2  (∵(a+b)2=a2+2ab+b2)  On further simplification,  f(x)=(x2+1)2−(2x)2  =(x2+1−2x)(x2+1+2x)(∵a2−b2=(a+b)(a−b))  Thus, the function f(x)=x4+1 into the product of irreducible quadratics is (x2+1−2x)(x2+1+2x)

Jenny Bolton · Accepted Answer

Step 1  Formula used:  The general quadratic equation is f(x)=ax2+bx+c  Here, x represents as unknow, while a, b and c are constants with a not equal to zero.  The values of x are x=−b±b2−4ac2a  Step 2  Calculation:  The given function is f(x)=x4+1  The product of irreducible quadratics is  (x2+1−2x)(x2+1+2x)  Use the above mentioned Formula and  solves the equation x2−2x+1 as follows,  Here, a=1  b=−2 and  c=1  Substitute a=1, b=12, and c=1 in x=−b±b2−4ac2a  x=−(−2)±(−2)2−4×1×12×1  =2±2−42  =2±−22  =2±2−12  =2±2i2 (∵i=−1)  Use the above mentioned Formula and solves the equation x2+2x+1 as follows,  Here, a=1, b=2 and c=1  Substitute a=1, b=2 and c=1 on x=−b±b2−4ac2a  x=−2±(2)2−4×1×12×1  =−2±2−42  =−2±−22  =−2±2−12  =−2±2i2 (∵i=−1)  Thus, the zeros of f(x)=x4

Determine: a) The function f(x)=x^{4}+1 into the product of irreducible quadratics. b)

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