Question

(2x²+3)²+(2x-3)² = (8x+6)(x-1)+22​

Answer 1

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Answer:

$({{2x}^{2}+3)}^{2}+{(2x-3)}^{2}= (8x+6)(x-1)+22$

$({16x}^{4} +9+{12x}^{2})+({4x}^{2}+9-12x)= (8x+6)(x-1)+22$

$(16{x}^{4}+9+{12x}^{2}+4{x}^{2}+9-12x)=8{x}^{2}-8x+6x-6+22$

$16{x}^{4}+8{x}^{2}+18-12x=8{x}^{2} -2x+16$

$16{x}^{4}+18-8{x}^{2} +2x-16=0$

$16{x}^{4}-8{x}^{2}+18+14x-16=0$

$16{x}^{4}-8{x}^{2}+14x+2=0$

$8{x}^{4}-8{x}^{2}+7x+1=0$

Hope it helps úh

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