Exercise:
Let g(x)= 4x4−4x3−71x2+36x+224
Show that , g(x) has 4 zeros to be determined
Solution: we have g(x)=4x4−4x3−71x2+36x+224=0
it’s clear that 224=315−91

⇒4x4−4x3−71x2+36x+315−91=0
⇒4x4−4x3−71x2+36x+315=91
To factorize p(x)=4x4−4x3−71x2+36x+315∈R[x]
So we need to find all divisors for 315 D315={1,3,5,7,9,15,21,35,45,63,105,315}
p(1)=4−4−71+36+315≠0 , p(3)=0 hence (x−3) is a factor of p(x)
So p(x)=(x−3)q(x) where deg(q(x))=3 and q(x)∈R[x]
Thus p(x)=(x−3)(ax3+bx2+cx+d)
Let’s expand p(x)=ax4+bx3+cx2+dx−3ax3−3bx2−3cx−3d
⇒p(x)=ax4+x3(b−3a)+x2(c−3b)+x(d−3c)−3d
Thus a=4,−3d=315,b−3a=−4,c−3b=−71,d−3c=36
So {a,b,c,d}={4,8,−47,−105}
Thus p(x)=(x−3)(4x3+8x2−47x−105)
Let’s find new factor for q(x)=4x3+8x2−47x−105
By the same miner we obtain after finding the divisors for 105 that (x+3) is factor of q(x)
So q(x)=(x+3)t(x) where t(x)∈R[x]°(t(x))=2
So expanding q(x)=(x+3)(ax2+bx+c)=ax3+bx2+cx+3ax2+3bx+3c
⇒q(x)=ax3+x2(b+3a)+x(c+3b)+3c
So a=4,b+3a=8,c+3b=−47&3c=−105 i.e {a,b,c}={4,−4,−35}
Hence p(x)=(x−3)(x+3)(4x2−4x−35)
But t(x)=4x2−4x−35=(2x)2−2(2x)(1)+1−1=35=(2x−1)2=36
So t(x)=(2x−1)2−36=(2x−1−6)(2x−1+6)=(2x−7)(2x+5)
Thus p(x)=(x−3)(x+3)(2x−7)(2x+5)
Therefore, (2x−7)(x+3)(x−3)(2x+5)=91
⇒(2x2−x−21)(2x2−x−15)=91 but −21=−15−6
⇒(2x2−x−15−6)(2x2−x−15)=91 put w=2x2−x−15
⇒(w−6)w=91⇒w2−6w−91=0⇒(w−3)2−9−91=(w−3)2−100=0
⇒(w−3−10)(w−3+10)=0⇒(w−13)(w+7)=0
Hence w=13orw=−7
ifw=13⇔2x2−x−15=13⇔2x2−x−28=0
ifw=−7⇔2x2−x−15=−7⇔2x2−x−8=0
2x2−x−28=x2−12x−14=x2−12x+116−116−14=(x−14)2−25516=0
⇒(x−14)2=25516⇒x−14=±√2554⇒x=1±√2554 (2 zeros )
Also 2x2−x−8=x2−12x−4=(x−14)2−116−4=(x−14)2−6516=0
So x−14=±√654⇔x=1±√654 ( 2 zeros )
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