Exercise 2.4

1. Determine which of the following polynomials has (x + 1) as a factor.

(i) x3 - x2 - x + 1

Solution:

YesNo

Substituting x = -1 gives −13 - −12 - (-1) + 1

= - + + =

(ii) x4 - x3 + x2 - x + 1

Solution:

NoYes

Substituting x = -1 gives + + + + =

(iii) x4 + 2x3 + 2x2 + x + 1

Solution:

NoYes

Substituting x = -1 gives - + - + =

(iv) x3 - x2 - 3−33x + 33

Solution:

NoYes

Substituting x = -1 gives + + + -3sqrt(3)3sqrt(3) + 3sqrt(3)sqrt(3) =

2. Use the Factor Theorem to determine whether g(x) is a factor of f(x) in each of the following cases :

(i) f(x) = 5x3 + x2 - 5x - 1, g(x) = x + 1

Solution:

YesNo

f(-1) = + + - 1 =

(ii) f(x) = x3 + 3x2 + 3x + 1, g(x) = x + 1

Solution:

YesNo

f(-1) = + - + 1 =

(iii) f(x) = x3 - 4x2 + x + 6, g(x) = x - 2

Solution:

YesNo

f(2) = - + + =

(iv) f(x) = 3x3 + x2 - 20x + 12, g(x) = 3x - 2

Solution:

YesNo

f23 = 8998 + 49−49 - 403203 + 12

=

(v) f(x) = 4x3 + 20x2 + 33x + 18, g(x) = 2x + 3

Solution:

YesNo

f−32 = −272272 + 452−452 - 992−992 + 18 =

3. Show that (x - 2), (x + 3) and (x - 4) are factors of x3 - 3x2 - 10x + 24.

Solution:

Let f(x) = x3 - 3x2 - 10x + 24.

f(2) = - - + 24 =

f(-3) = - + + 24 =

f(4) = - - + 24 =

Since f(2) = f(-3) = f(4) = , (x-2), (x+3), and (x-4) are of f(x).

4. Show that (x + 4), (x - 3) and (x - 7) are factors of x3 - 6x2 - 19x + 84.

Solution:

Let f(x) = x3 - 6x2 - 19x + 84

f(-4) = - + + 84 =

f(3) = - - + 84 =

f(7) = - - + 84 =

Since f(-4) = f(3) = f(7) = , (x+4), (x-3), and (x-7) are of f(x).

5. If both (x - 2) and (12 x - 1) are factors of px2 + 5x + r, then show that p = r.

Solution:

Let f(x) = px2 + 5x + r.

Since (x-2) is a factor, f(2) = 4p4p2 + + r = .

Since (12x−1) or (x-2) is a factor, f(2) = 4p + 10 + r = .

So, 4p + 10 + r = 0 => r = p - .

Since (x-2) is a factor, we can write f(x) = p(x-2)(x-2) = p(x2−x2 - 4x2x + 42) = px2 - + .

Comparing coefficients, 5 = and r = .

So, p = −5454 and r = 4(−5454) = .

This doesn't show p = r.

If (x-2) and (2x-2) are factors, then f(x) = p(x-2)(2x-22x+2) = p(2x2x2 - x + ) = 2px2 - x +

So, 5 = => p = −5656 and r = 4p = −206206 = −103103.

If (x-2) and (x-2) are factors, then p = and r = .

If (x-2) and (2x-2) are factors, then p = and r = .

If (x-2) and (x-2) are factors, then p = r = .

6. If x2−1 is a factor of ax4 + bx3 + cx2 + dx + e, then show that a + c + e = b + d = 0.

Solution:

Since x2−1 = (x-1)(x+1) is a factor, x = and x = -1 are roots.

f(1) = + + + + e = 0

f(-1) = + + + + e = 0

Adding these equations: + + = 0 => + + = 0

Subtracting these equations: + = 0 => + = 0

Therefore, a + c + e = b + d =

7. Factorise

(i) x3 - 2x2 - x + 2

Solution:

x2( - ) -+ (x-2)

= (x2-1)(x-2)

= (x-1)(x+1x-1)(x-2)

(ii) x3 - 3x2 - 9x - 5

Solution:

Given polynomial equation: x3 - 3x2 9x - 5.

By Trial and Error mwthod

Let, f(x) = x3 - 3x2 - 9x - 5

f(1) = 13 - 312 - 9(1) - 5

= - - - =

f(1) = −13 - 3−12 - 9(-1) - 5

- + - =

Therefore, (x+1) is a factor of f(x).

By dividing f(x) with (x+1) we get (x2-4x-5) as another factor.

(x+1)(x2-4x-5)

= (x+1)(x-5x+5)(x+1)

= (x+12x−12) (x-5)

(iii) x3 + 13x2 + 32x + 20

Solution:

x3 + x2 + 12x2 + 12x + 20x + 20

= x2(x+1x-1) + 12x12x2 (x+1) + (x+1)

= (x+1)(x2 + 12x + 20)

= (x+1)(x+2x-2)(x+10x+5)

(iv) y3 + y2 - y - 1

Solution:

y2y+1 - (y+1) = (y2-1)(y+1)

= (y-1)(y+1y-1)(y+1)

= y−1y+12y+1y−12

8. If ax2 + bx + c and bx2 + ax + c have a common factor x + 1 then show that c = 0 and a = b.

Solution:

If (x+1) is a factor of ax2 + bx + c, then a−12 + b(-1) + c = 0, so - + c = 0.

If (x+1) is a factor of bx2 + ax + c, then b−12 + a(-1) + c = 0, so - + c = 0.

Adding the two equations: (a - b + c) + (b - a + c) = 0 + 0, which simplifies to = 0, so c = .

Substituting c = 0 into a - b + c = 0 gives a - b = 0, so = b.

9. If x2 - x - 6 and x2 + 3x - 18 have a common factor (x - a) then find the value of a.

Solution:

x2 - x - 6 = (x-3x+3)(x+2)

x2 + 3x - 18 = (x+6x+3)(x-3)

The common factor is (x-3), so a = .

10. If (y - 3) is a factor of y3 - 2y2 - 9y + 18 then find the other two factors.

Solution:

y3 - 2y2 - 9y + 18 = y2(y-2y+2) - (y-2)

= y2−9y−2

= y−3y+3y−2y+32y−2

The other two factors are (y+3y-3) and (y-2y+2).