Suggested: one of the factors of 2x^3+42x^2+208x is x+b - 7(y+3)-2x+2)=14 4(y-2)+3(x-3)=2 by substitution method - 3/4(7x-1)-(2x-1-x/2)=x+3/2 - x+1/x-1+x-2/x+2=4-2x+3/x-2 - if x+1 is a factor of 2x^3+ax^2+2bx+1 and 2a-3b=4 - (3x^2y^4+2xy)dx+(2x^3y^3-x^2)dy=0 - 2^2x-2^x+3+2^4=0 - solve 2x^5+x^4-12x^3-12x^2+x+2=0 - gauss jordan method 2x+3y-z=5 4x+4y-3z=3 2x-3y+2z=2 - gauss elimination method 2x+3y-z=5 4x+4y-3z=3 2x-3y+2z=2 - 2x^2-x-1 4x^2+8x+3 hcf - 2x^2-4x+3=0 quadratic equation - 6(2x^2-3x-2) 8(4x^2+4x+1) 1 2(2x^2+7x+3) - 1+2x+3x^2+4x^3 to infinity - 4x^3+2x^2 Browse related:
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