**Find the remainder when x**^{3}+ 3x^{2}+ 3x + 1 is divided by

**(i) x + 1**

**(ii) x – ^{1}/_{2}**

**(iii) x**

**(iv) x + π**

**(v) 5 + 2x**

Solution:

(i) x + 1

p(x) = x^{3} + 3x^{2} + 3x + 1 and the zero of x + 1 is – 1

So, p(1) = (-1)^{3} + 3(-1)^{2} + 3(-1) + 1

= -1 + 3 – 3 + 1

= 0

So, by the remainder theorem, remainder obtained on dividing p(x) by x + 1 is 0.

So, p(x) is the multiple of x + 1

(ii) x – ^{1}/_{2}

p(x) = x^{3} + 3x^{2} + 3x + 1 and zero of x – ^{1}/_{2 }is ^{1}/_{2}

p(^{1}/_{2}) = (^{1}/_{2})^{3} + 3(^{1}/_{2})^{2} + 3(^{1}/_{2}) + 1

= ^{1}/_{8} + 3*^{1}/_{4} + 3*^{1}/_{2 }+ 1

= ^{1}/_{8} + ^{3}/_{4 }+ ^{3}/_{2} + 1

= ^{1+6+12+8}/_{8} = ^{27}/_{8}

So, by the remainder theorem, remainder obtained on dividing p(x) by x – ^{1}/_{2} is ^{27}/_{8}.

(iii) x

p(x) = x^{3} + 3x^{2} + 3x + 1 and zero of x _{ }is 0

p(0) = (0)^{3} + 3(0)^{2} + 3(0) + 1

= 0 + 3*0 + 3*0 + 1

= 1

So, by the remainder theorem, remainder obtained on dividing p(x) by x is 1.

(iv) x + π

p(x) = x^{3} + 3x^{2} + 3x + 1 and zero of x + π is -π

p(-π) = (-π)^{3} + 3(-π)^{2} + 3(-π) + 1

= -π^{3} + 3π^{2} – 3π + 1

So, by the remainder theorem, remainder obtained on dividing p(x) by x + π is = -π^{3} + 3π^{2} – 3π + 1.

(v) 5 + 2x

p(x) = x^{3} + 3x^{2} + 3x + 1 and zero of 5 + 2x _{ }is –^{5}/_{2}

p(-^{5}/_{2}) = (-^{5}/_{2})^{3} + 3(-^{5}/_{2})^{2} + 3(-^{5}/_{2}) + 1

= –^{125}/_{8} +3*^{25}/_{4} – 3*^{5}/_{2} + 1

= ^{-125+150-60+8}/_{8}

= ‑^{27}/_{8}

So, by the remainder theorem, remainder obtained on dividing p(x) by 5 + 2x is –^{27}/_{8}.

- Find the remainder when x
^{3}– ax^{2}+ 6x – a is divided by x – a

Solution:

p(x) = x^{3} – ax^{2} + 6x – a and zero of x – a is a

p(1) = a^{3} – a(a)^{2} + 6(a) – a

= a^{3} – a^{3} + 6a – a

= 5a

So, by the remainder theorem, remainder obtained on dividing p(x) by x – a is 5a.

- Check whether 7 + 3x is a factor of 3x
^{3}+ 7x

Solution:

7 + 3x = 0

7 = – 3x

x = –^{7}/_{3}

Zero of 7 + 3x is –^{7}/_{3}

p(-^{7}/_{3}) = 3x^{3} + 7x

= 3(-^{7}/_{3})^{3} + 7(-^{7}/_{3})

= 3(-^{343}/_{27}) – 7(^{7}/_{3})

= – ^{343}/_{9} – ^{49}/_{3}

= –^{490}/_{9}

So, remainder = – ^{490}/_{9} which is different from 0.

Therefore, (3x + 7) is not a factor of the polynomial 3x^{3} + 7x

## 2 thoughts on “Polynomial – Exercise 2.3 – Class IX”

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