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**Question 1. Show that (x + 3) is a factor of x ^{3} + x^{2} – 4x + 6.** ∵ p(x) = x

Solution:

Since (x + 3) is a factor, then x + 3 = 0 ⇒ x = – 3

∴ p( – 3) = ( – 3)

= – 27 + 9 + 12 + 6

= 0

∴ (x + 3) is a factor of x

**Question 2. Show that (x – 5) is a factor of: x ^{3} – 3x^{2 } – 13x + 15** ∵ p(x) = x

Solution:

Since (x – 5) is a factor, then x – 5 = 0

⇒ x = 5

∴ p(5) = (5)

= 125 – 75 – 65 + 15 = 140 – 140 = 0

∴ (x – 5) is a factor of x

**Question 3. Find the value of a such that (x + α ) is a factor of the polynomial f(x) = x ^{4} – α 2x^{2} + 2x + α + 3. Solution:** Here f(x) = x

Since, (x + a) is a factor of f(x)

∴ f ( –

⇒ ( –

⇒

⇒ –

⇒

**Question 4. Show that (x – 2) is a factor of 3x ^{3} + x^{2} – 20x + 12.** f(x) = 3x

Solution:

For (x – 2) being a factor of f(x), then x – 2 = 0

⇒ x = 2

∴ f(2) must be zero.

Since f(2) = 3(2)

= 3(8) + 4 – 40 + 12

= 24 + 4 – 40 + 12

= 40 – 40 = 0

which proves that (x – 2) is a factor of f(x).

**Question 5. Factorise the polynomial **

**Solution**: We have

**Question 6. Factorise a(a – 1) – b(b – 1). Solution:** We have a(a – 1) – b(b – 1)

= a

= a

(Rearranging the terms) = [(a + b)(a – b)] – (a – b)

[∵ x

= (a – b)[(a + b) – 1]

= (a – b) (a + b – 1)

Thus, a(a – b) – b(b – 1) = (a – b)(a + b – 1)

Solution:

∴ x

= [(x + y)(x + y)

= (x + y)[(x + y)

= (x + y)[(x

= (x + y)[x

= (x + y)[x

Thus, x

Solution

∴ x

= [(x – y)(x – y)

= (x – y)[(x – y)

= (x – y)[(x

= (x – y)[x

= (x – y)(x

Thus, x

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