Question

Factorise the followi
4x4 + 8x² +9
a.​

Answer 1

Answer:

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Step-by-step explanation:

polynomial methods, by considering  x2  as the indeterminate.

(Or by substituting  x2  by  t , if you find that easier to read.)

Since  4t2+8t+9  has no real roots (16–36 is negative), this quadratic cannot be factored into two linear factors.

If you want to factor over the complex numbers:

4t2+8t+9=(2t+2+5–√i)(2t+2−5–√i)  

so the original biquadratic polynomial factors as

4x4+8x2+9=(2x2+2+5–√i)(2x2+2−5–√i)  

which further factors (over the complex numbers, that is) into the four linear terms

x  ±12±125–√i .

Recombining those two by two such that the imaginary part cancels out, finally gives the following “real” factorisation:

4x4+8x2+9=(2x2+2x+3)(2x2−2x+3)  

In general, a biquadratic polynomial  px4+qx2+r  which does not factor as  a(x2−b)(x2−c)  for some real numbers  a,b,c  will always factor as  (ax2+bx+c)(ax2−bx+c)  for some real numbers  a,b,c :

(1) possibly change signs of  p,q,r  if  p  is negative; (2) set  a=q√ ; set  c=r√ ; (3) set  b=2ac−q−−−−−−√ .

Only when this  2ac−q  is positive, you have a “real” factorisation; otherwise, both  x  coefficients are pure imaginary.