# EQUATING COEFIFICIENTS OF POLYNOMIALS WORKSHEET

Problem 1 :

Find constants a, b and c given that

2x2 + 4x + 5 = ax2 + [2b - 6]x + c for all x

Solution

Problem 2 :

2x3 - x2 + 6 = (x - 1)2 (2x + a) + bx + c for all x.

Solution

Find a and b if :

Problem 3 :

z4 + 4 = (z2 + az + 2) (z2 + bz + 2) for all z.

Solution

Problem 4 :

2z4 + 5z3 + 4z2 + 7z + 6 = (z2 + az + 2) (2z2 + bz + 3) for all z.

Solution

Problem 5 :

Show that z4 + 64 can be factorised into two real quadratic factors of the form z2 + az + 8 and z2 + bz + 8, but cannot be factorised into two real quadratic factors of the form z2 + az + 16 and z2 + bz + 4.

Solution

Problem 6 :

Find real numbers a and b such that

x4 - 4x2 - 8x - 4 = (x2 + ax + 2) (x2 + bx - 2)

and hence solve the equation. x4 + 8x = 4x2 + 4.

Solution

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 2z3 – z2 + az – 3. Find a and all zeros of the cubic.

Solution

Problem 2 :

3z + 2 is a factor of 3z3 – z2 + [a + 1]z + a. Find a and all the zeros of the cubic.

Solution

Problem 3 :

Both 2x + 1 and x – 2 are factors of P(x) = 2x4 + ax3 + bx2 – 12x – 8. Find a and b and all zeros of P(x).

Solution

Problem 4 :

x + 3 and 2x – 1 are factors of 2x4 + ax3 + bx2 + ax + 3. Find a and b and hence determine all zeros of the quartic.

Solution

Problem 5 :

a. x3 + 3x2  - 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.

Solution

Problem 6 :

3x3 + 4x2 – x + m has two identical linear factors. Find m and find the zeros of the polynomial in all possible cases.

Solution

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).

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