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Algebra Exercise asked in the math group and edited by me in different way


Exercise:

Let g(x)= 4x44x371x2+36x+224

Show that , g(x)  has 4 zeros to be determined

Solution: we have g(x)=4x44x371x2+36x+224=0

it’s clear that 224=31591

4x44x371x2+36x+31591=0

4x44x371x2+36x+315=91

To factorize p(x)=4x44x371x2+36x+315R[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)=4471+36+3150 , p(3)=0 hence (x3) is a factor of p(x)

So p(x)=(x3)q(x) where deg(q(x))=3 and q(x)R[x]

Thus p(x)=(x3)(ax3+bx2+cx+d)

Let’s expand p(x)=ax4+bx3+cx2+dx3ax33bx23cx3d

p(x)=ax4+x3(b3a)+x2(c3b)+x(d3c)3d

Thus a=4,3d=315,b3a=4,c3b=71,d3c=36

So {a,b,c,d}={4,8,47,105}

Thus p(x)=(x3)(4x3+8x247x105)

Let’s find new factor for q(x)=4x3+8x247x105

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]&deg(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)=(x3)(x+3)(4x24x35)

But t(x)=4x24x35=(2x)22(2x)(1)+11=35=(2x1)2=36

So t(x)=(2x1)236=(2x16)(2x1+6)=(2x7)(2x+5)

Thus p(x)=(x3)(x+3)(2x7)(2x+5)

Therefore, (2x7)(x+3)(x3)(2x+5)=91

(2x2x21)(2x2x15)=91 but 21=156

(2x2x156)(2x2x15)=91 put w=2x2x15

(w6)w=91w26w91=0(w3)2991=(w3)2100=0

(w310)(w3+10)=0(w13)(w+7)=0

Hence w=13orw=7

ifw=132x2x15=132x2x28=0

ifw=72x2x15=72x2x8=0

2x2x28=x212x14=x212x+11611614=(x14)225516=0

(x14)2=25516x14=±2554x=1±2554  (2 zeros )

Also 2x2x8=x212x4=(x14)21164=(x14)26516=0

So x14=±654x=1±654 ( 2 zeros )

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