Question

Find zeros of the polynomial 2 x power 4 - 10 x cube + 5 x square + 15 x minus 12

Answer 1

the polynomial p(x)=2x^4-10x^3+5x^2+15x-12
now substitute x=1 in polynomial p(x)
I.e., p(x)=2(1)^4-10(1)^3+5(1)^2+15(1)-12
=2(1)-10(1)+5(1)+15(1)-12
=2-10+5+15-12
=22-22
=0
therefore 1 is the zero of the polynomial p(x)
therefore x-1 is the factor of the polynomial p(x)
now substitute x=4 in the polynomial p(x)
I.e., p(x)=2(4)^4-10(4)^3+5(4)^2+15(4)-12
=2(256)-10(64)+5(16)+15(4)-12
=512-640+80+60-12
=652-652
=0
therefore 4 is also zero of the polynomial p(x)
therefore x-4 is also factor of the polynomial p(x)
now multiply both the factors
I.e., (x-1)(x-4)=x(x-4)-1(x-4)
=x^2-4x-x+4
=x^2-5x+4
therefore x^2-5x+4 is also the factor of the polynomial p(x)
now divide 2x^4-10x^3+5x^2+15x-12 by x^2-5x+4
I.e., x^2-5x+4)2x^4-10x^3+5x^2+15x-12(2x^2-3
2x^4-10x^3+8x^2
-___+____-__________
-3x^2+15x-12
-3x^2+15x-12
+___-___+__
____0____
therefore 2x^2-3 is also a factor of the polynomial p(x)
now the quadratic polynomial is 2x^2-3
now compare the quadratic polynomial by the polynomial ax^2+bx+c
I.e., a=2,b=0,c=-3
quadratic formula,
=[-b+√(b^2-4ac)]/2a, [-b-√(b^2-4ac)]/2a
=[-0+√{0-4*2*(-3)}]/2*2, [-0-√{0-4*2(-3)}]/2*2
=0+(√24)/4, 0-(√24)/4
=(√24)/4, -(√24)/4
=2(√6)/4, -2(√6)/4
=(√6)/2, -(√6)/2
therefore zeroes of the polynomial p(x) are 1,4,(√6)/2 and -(√6)/2