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We will use the formula A^2-B^2=(A-B)(A+B)

28 March 2013

So (x+c)^4 - (a-c)^4=[(x+c)^2- (a-c)^2][(x+c)^2 + (a-c)^2]=[(x^2+2xc+c^2)-(a^2-2ac+c^2)][(x^2+2xc+c^2)+(a^2-2ac+c^2)]=(x^2+2xc+c^2-a^2+2ac-c^2)(x^2+2xc+c^2+a^2-2ac+c^2)=(x^2-a^2+2xc+2ac)(x^2+2xc+a^2-2ac+2c^2)= (x-a)(x+a)+2c(x+a)=(x+a)(x-a+2c)(x^2+2xc+a^2-2ac+2c^2) so (x+a)^4 + (x+c)^4 - (a-c)^4=(x+a)^4+(x+a)(x-a+2c)(x^2+2xc+a^2-2ac+2c^2) =(x+a)[(x+a)^3+(x-a+2c)(x^2+2xc+a^2-2ac+2c^2)], therefore x+a is a factor of the polynomial. In this demonstration we have also used the formula (A+B)^2=A^2+2AB+B^2 and (A-B)^2=A^2-2AB+B^2

28 March 2013

Please read this demonstration as there were few corrections. Thank you. So (x+c)^4 - (a-c)^4=[(x+c)^2- (a-c)^2][(x+c)^2 + (a-c)^2]=[(x^2+2xc+c^2)-(a^2-2ac+c^2)][(x^2+2xc+c^2)+(a^2-2ac+c^2)]=(x^2+2xc+c^2-a^2+2ac-c^2)(x^2+2xc+c^2+a^2-2ac+c^2)=(x^2-a^2+2xc+2ac)(x^2+2xc+a^2-2ac+2c^2)= (x-a)(x+a)+2c(x+a)=(x+a)(x-a+2c)(x^2+2xc+a^2-2ac+2c^2) so (x+a)^4 + (x+c)^4 - (a-c)^4=(x+a)^4+(x+a)(x-a+2c)(x^2+2xc+a^2-2ac+2c^2) =(x+a)[(x+a)^3+(x-a+2c)(x^2+2xc+a^2-2ac+2c^2)], therefore x+a is a factor of the polynomial. In this demonstration we have also used the formula (A+B)^2=A^2+2AB+B^2 and (A-B)^2=A^2-2AB+B^2

28 March 2013

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