Problem 1 :
Find constants a, b and c given that
2x^{2} + 4x + 5 = ax^{2} + [2b - 6]x + c for all x
Problem 2 :
2x^{3} - x^{2} + 6 = (x - 1)^{2} (2x + a) + bx + c for all x.
Find a and b if :
Problem 3 :
z^{4} + 4 = (z^{2} + az + 2) (z^{2} + bz + 2) for all z.
Problem 4 :
2z^{4} + 5z^{3} + 4z^{2} + 7z + 6 = (z^{2} + az + 2) (2z^{2} + bz + 3) for all z.
Problem 5 :
Show that z^{4} + 64 can be factorised into two real quadratic factors of the form z^{2} + az + 8 and z^{2} + bz + 8, but cannot be factorised into two real quadratic factors of the form z^{2} + az + 16 and z^{2} + bz + 4.
Problem 6 :
Find real numbers a and b such that
x^{4} - 4x^{2} - 8x - 4 = (x^{2} + ax + 2) (x^{2} + bx - 2)
and hence solve the equation. x^{4} + 8x = 4x^{2} + 4.
1) So, the values of a, b and c is 2, 5 and 5.
2) So, the values of a, b and c is 3, 4 and 3.
3) So, the values of a and b is -2 and 2.
4) a = 3, b = -1
6) a = 1, b = 2 and c = -2 and z = -1 ±√3
Problem 1 :
2z – 3 is a factor of 2z^{3} – z^{2} + az – 3. Find a and all zeros of the cubic.
Problem 2 :
3z + 2 is a factor of 3z^{3} – z^{2} + [a + 1]z + a. Find a and all the zeros of the cubic.
Problem 3 :
Both 2x + 1 and x – 2 are factors of P(x) = 2x^{4} + ax^{3} + bx^{2} – 12x – 8. Find a and b and all zeros of P(x).
Problem 4 :
x + 3 and 2x – 1 are factors of 2x^{4} + ax^{3} + bx^{2} + ax + 3. Find a and b and hence determine all zeros of the quartic.
Problem 5 :
a. x^{3} + 3x^{2} - 9x + c has two identical linear factors. Prove that c is either 5 or -27 and factories the cubic into linear factors in each case.
Problem 6 :
3x^{3} + 4x^{2} – x + m has two identical linear factors. Find m and find the zeros of the polynomial in all possible cases.
1) a = -1, zeroes are 3/2 and (-1±,i√3)/2
2) a = 6, zeroes are -2/3 and (1±,i√11)/2
3) The zeroes are x = ±i2, x = -1/2 x = 2
4) a = -11 and b = 15
5) P(x) = (x - 3) (x + 3)^{2}.
6) repeated factor is (x + 1).
May 21, 24 08:51 PM
May 21, 24 08:51 AM
May 20, 24 10:45 PM