1) For the function of&nbsp;f(x)=x4+1, factor f into the product of two irreducible quadratics. (Hint: complete the square by adding and subtracting&nbsp;2x2)&nbsp;2) Find the zeros of f by finding the zeros of each irreducible quadratic.

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1) For the function of&amp;nbsp;f(x)=x4+1, factor f into the product of two irreducible quadratics. (Hint: complete the square by adding and subtracting&amp;nbsp;2x2)&amp;nbsp;2) Find the zeros of f by finding the zeros of each irreducible quadratic.

SlabydouluS62 · Accepted Answer

Step 11) For a fuction of&amp;nbsp;f(x)=x4+1, factor f into the product of two Irreducible quadratics2) To determine the zeros of each irreducible quadratic before moving on to f.Step 2Given:&amp;nbsp;f(x)=x4+1Using completing the square by adding and subtracting&amp;nbsp;2x2, we getf(x)=x4+1+2x2−2x2writin asf(x)=(x2)2+2x2+(1)2−2x2⇒f(x)=(x2+1)2−2x2(∵a2+2ab+b2=(a+b)2&amp;nbsp;here&amp;nbsp;a=x2b=1Now using&amp;nbsp;l2−m2=(l−m)(l−m)We have here&amp;nbsp;l=x2+1m=2xi.e&amp;nbsp;f(x)=(x2+1)2−(2x)2⇒f(x)=(x2+1−2x)(x2+1+2x)⇒f(x)=(x2−2x+1)(x2+2x+1)There are the factors of&amp;nbsp;f(x)Step 3Two irreducible quadratics are presentLetg(x)=x2−2x+1h(x)=x2+2x+1∴f(x)=g(x)×h(x)Now To find zeros of fFirstly finding zeros of g(x) and h(x)Put&amp;nbsp;g(x)=0⇒x2−2x+1=0Comparring with&amp;nbsp;ax2+bx+c=0⇒x=−b±b2−4ac2aUsing this we have&amp;nbsp;

Maricela Alarcon · Accepted Answer

Step 1  f(x)x4+1  a) f(x)=x4+2x2+1−2x2  =(x2+1)2−(2x)2  Using (a2−b2)=(a+b)(a−b)  f(x)=(x2+1+2x)(x2+1−2x)  b) Zeroes x2+2x+1=0, x2−2x+1=0  Using quadratic formula which is  x=−b±b2−4ac2a  x=−2±2−4(1)(1)2  =−2±i22, 2±i22  No real zeroes complex zeroes are  −2±22, 2±i22

Step 1 a) For the function of f(x)=x^{4}+1, factor f into

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