1. find all the4 zeroes of the polynomial 2x^4 + 7x^3 - 19x^2 - 14x +30, if two of its zeroes are root 2 and -root 2
2. find the zeroes of the quadratic polynomial root3x^@ - 8x + 4root3
3. find all the zeroes of the polynomial 2x^3 - 4x - x^2 +2, if two of its zeroes are root2 ans -root2
4. if the polynomial 6x^4 + 8x^3 [email protected] +21x +7 is divide by another polynomial 3x^2 + 4x +1, the remainder comes out to be ax+b, find 'a' and 'b' .

Answer in 4 parts:

1. (x-√2)(x+√2)=x^2-2, so divide x^2-2 into 2x^4+7x^3-19x^2-14x+30 using long division: 2x^2+7x-15 which factorises: (2x-3)(x+5), so other roots are 3/2 and -5.
2. Use the formula to find the roots: x=(8±√(64-48))/2√3=(8±4)/2√3=6/√3 and 2/√3; rationalised these zeroes are 6√3/3=2√3 and 2√3/3.
3. The polynomial can be rewritten: 2x(x^2-2)-(x^2-2)=(2x-1)(x^2-2), so the third zero is 1/2.
4. The quotient starts with 2x^2 because 2x^2 * 3x^2=6x^4. The whole multiple of the divisor becomes: 6x^4+8x^3+2x^2. Subtracting this from the dividend we get: 15x^2+21x+7. 5 times the divisor gives us: 15x^2+20x+5, and subtracting this from the previous remainder we get: x+2 so a=1 and b=2.
by Top Rated User (764k points)