Chapter 2: Polynomials

Ans: We have,

 $ p\left( z \right) = a{z^3} + 4{z^2} + 3z - 4 $ 

And  $ q\left( z \right) = {z^3} - 4z + a $ 

As it is given that both the Polynomials leave the same remainder when divided by  $ z - 3 $ .

Therefore,

 $ p\left( 3 \right) = q\left( 3 \right) $ 

\[ \Rightarrow a{\left( 3 \right)^3} + 4{\left( 3 \right)^2} + 3\left( 3 \right) - 4 = \,{\left( 3 \right)^3} - 4\left( 3 \right) + a\]

\[ \Rightarrow 27a + 4 \times 9 + 9 - 4 = \,27 - 12 + a\]

\[ \Rightarrow 27a + 36 + 5 = \,15 + a\]

 $  \Rightarrow 27a - a = 15 - 41 $ 

 $  \Rightarrow 26a =  - 26 $ 

 $  \Rightarrow a =  - 1 $ 

Hence, the required value of a is  $  - 1 $ .

Ans: We know that when  $ p\left( x \right) $  is divided by  $ x + b $ , then the remainder  $  = \,p\left( { - b} \right) $ 

Now,  $ p\left( x \right) = {x^4} - 2{x^3} + 3{x^2} - ax + 3a - 7 $ is divided by  $ x + 1 $ , then the remainder  $  = \,p\left( { - 1} \right) $ .

Also, we know that the remainder is  $ 19 $ .

Therefore,

 $ \,p\left( { - 1} \right) = 19 $ 

 $ {\left( { - 1} \right)^4} - 2{\left( { - 1} \right)^3} + 3{\left( { - 1} \right)^2} - a\left( { - 1} \right) + 3a - 7 = 19 $ 

 $  \Rightarrow 1 + 2 + 3 + a + 3a - 7 = 19 $ 

 $  \Rightarrow 4a - 1 = 19 $ 

 $  \Rightarrow 4a = 20 $ 

 $  \Rightarrow a = 5 $ 

Now,  $ p\left( x \right) = {x^4} - 2{x^3} + 3{x^2} - 5x + 3\left( 5 \right) - 7 = {x^4} - 2{x^3} + 3{x^2} - 5x + 8 $ 

Again, when p(x) is divided by  $ x + 2 $ , then

Remainder  $  = p\left( { - 2} \right) = {\left( { - 2} \right)^4} - 2{\left( { - 2} \right)^3} + 3{\left( { - 2} \right)^2} - 5\left( { - 2} \right) + 8 $ 

 $  = 16 + 16 + 12 + 10 + 8 $ 

 $  = 62 $ 

Ans: Let  $ q\left( x \right) = p{x^2} + 5x + r $ 

As  $ \left( {x - 2} \right) $  is a factor of  $ q\left( x \right) $ 

Then,  $ q\left( 2 \right) = 0 $ 

 $  \Rightarrow p{\left( 2 \right)^2} + 5\left( 2 \right) + r = 0 $ 

 $  \Rightarrow 4p + 10 + r = 0 $ 

 $  \Rightarrow  - 4p - r = 10............\left( 1 \right) $ 

Again,  $ \left( {x - \dfrac{1}{2}} \right) $  is a factor of  $ q\left( x \right) $ .

Then,  $ q\left( {\dfrac{1}{2}} \right) = 0 $ 

 $  \Rightarrow p{\left( {\dfrac{1}{2}} \right)^2} + 5\left( {\dfrac{1}{2}} \right) + r = 0 $ 

 $  \Rightarrow \dfrac{p}{4} + \dfrac{5}{2} + r = 0 $ 

 $  \Rightarrow p + 10 + 4r = 0 $ 

Now, using equation  $ 1 $ 

 $  \Rightarrow p - 4p - r + 4r = 0 $ 

 $  \Rightarrow  - 3p + 3r = 0 $ 

 $  \Rightarrow p = r $ 

Hence, proved.

Ans: We have,

 $ {x^2} - 3x + 2 $ 

We can also write it as

 $  \Rightarrow {x^2} - 3x + 2 = {x^2} - 2x - x + 2 $ 

 $  \Rightarrow x\left( {x - 2} \right) - 1\left( {x - 2} \right) = \left( {x - 1} \right)\left( {x - 2} \right) $ 

Now, we have

 $ p\left( x \right) = 2{x^4} - 5{x^3} + 2{x^2} - x + 2 $ 

Put  $ x = 1 $ 

 $ p\left( 1 \right) = 2{\left( 1 \right)^4} - 5{\left( 1 \right)^3} + 2{\left( 1 \right)^2} - \left( 1 \right) + 2 $ 

 $ p\left( 1 \right) = 2 - 5 + 2 + 1 = 0 $ 

 $ p\left( 1 \right) = 0 $ 

Therefore,  $ x - 1 $  divides p(x).

Put  $ x = 2 $ 

 $ p\left( 2 \right) = 2{\left( 2 \right)^4} - 5{\left( 2 \right)^3} + 2{\left( 2 \right)^2} - \left( 2 \right) + 2 $ 

 $ p\left( 2 \right) = 32 - 40 + 8 - 2 + 2 = 0 $ 

 $ p\left( 2 \right) = 0 $ 

Therefore,  $ x - 2 $  divides p(x).

Hence, $ {x^2} - 3x + 2 $ divides $ 2{x^4} - 5{x^3} + 2{x^2} - x + 2 $ .

Ans: We have,

 $  = {\left( {2x - 5y} \right)^3} - {\left( {2x + 5y} \right)^3} $ 

Now, using  $ {a^3} - {b^3} = \left( {a - b} \right)\left( {{a^2} + ab + {b^2}} \right) $ 

 $  = \left\{ {\left( {2x - 5y} \right) - \left( {2x + 5y} \right)} \right\}\left\{ {{{\left( {2x - 5y} \right)}^2} + \left( {2x - 5y} \right)\left( {2x + 5y} \right) + {{\left( {2x + 5} \right)}^2}} \right\} $ 

 $  = \left( { - 10y} \right)\left( {2{x^2} + 25{y^2}} \right) $ 

 $  =  - 120{x^2}y - 250{y^3} $ 

Ans: We have,

 $ \left( { - z + x - 2y} \right)\left( {{x^2} + 4{y^2} + {z^2} + 2xy + xz - 2yz} \right) $ 

 $  = \left\{ {x + \left( { - 2y} \right) + \left( { - z} \right)} \right\}\left\{ {{{\left( x \right)}^2} + {{\left( { - 2y} \right)}^2} + {{\left( { - z} \right)}^2} - \left( x \right)\left( { - 2y} \right) - \left( x \right)\left( { - z} \right) - \left( -2y \right)\left( { - z} \right)} \right\} $ 

 $  = {x^3} + {\left( { - 2y} \right)^3} + {\left( { - z} \right)^3} - 3\left( x \right)\left( { - 2y} \right)\left( { - z} \right) $ 

 $  = {x^3} - 8{y^3} - {z^3} - 6xyz $ 

Ans: Here, a, b, c are all non-zero and \[a + b + c = 0\].

Therefore,

 $ {a^3} + {b^3} + {c^3} = 3abc $ 

Now, we will divide the complete equation by abc.

 $ \dfrac{{{a^3}}}{{abc}} + \dfrac{{{b^3}}}{{abc}} + \dfrac{{{c^3}}}{{abc}} = \dfrac{{3abc}}{{abc}} $ 

 $ \dfrac{{{a^2}}}{{bc}} + \dfrac{{{b^2}}}{{ca}} + \dfrac{{{c^2}}}{{ab}} = 3 $ .

Ans: We know that, 

 $ {a^3} + {b^3} + {c^3} - 3abc = \left( {a + b + c} \right)\left( {{a^2} + {b^2} + {c^2} - ab - bc - ca} \right) $ 

 $  = \left( {a + b + c} \right)\left( {{a^2} + {b^2} + {c^2} - \left( {ab + bc + ca} \right)} \right) $ 

It is given that  $ a + b + c = 5 $  and  $ ab + bc + ca = 10 $ .

 $  = 5\left( {{a^2} + {b^2} + {c^2} - 10} \right) $ 

Now, we have 

 $ a + b + c = 5 $ 

Squaring both sides, we get

 $ {\left( {a + b + c} \right)^2} = {\left( 5 \right)^2} $ 

 $ {a^2} + {b^2} + {c^2} + 2\left( {ab + bc + ca} \right) = 25 $ 

 $ {a^2} + {b^2} + {c^2} + 2\left( {10} \right) = 25 $ 

 $ {a^2} + {b^2} + {c^2} = 25 - 20 = 5 $ 

Now, put the above value in equation  $  = 5\left( {{a^2} + {b^2} + {c^2} - 10} \right) $ 

 $  = 5\left( {5 - 10} \right) $ 

 $  = 5 \times  - 5 =  - 25 $ 

Hence, proved.

Ans:  $ {\left( {a + b + c} \right)^3} = {\left[ {a + \left( {b + c} \right)} \right]^3} $ 

 $  = {a^3} + 3{a^2}\left( {b + c} \right) + 3a{\left( {b + c} \right)^2} + {\left( {b + c} \right)^3} $ 

 $  = {a^3} + 3{a^2}b + 3{a^2}c + 3a\left( {{b^2} + 2bc + {c^2}} \right) + \left( {{b^3} + 3{b^2}c + 3b{c^2} + {c^3}} \right) $ 

 $  = {a^3} + 3{a^2}b + 3{a^2}c + 3a{b^2} + 6abc + 3a{c^2} + {b^3} + 3{b^2}c + 3b{c^2} + {c^3} $ 

 $  = {a^3} + {b^3} + {c^3} + 3{a^2}b + 3{a^2}c + 3a{b^2} + 6abc + 3a{c^2} + 3{b^2}c + 3b{c^2} $ 

We can also write it as,

 $  = {a^3} + {b^3} + {c^3} + 3abc + 3{a^2}b + 3a{c^2} + 3{a^2}c + 3{b^2}c + 3{b^2}a + 3b{c^2} + 3abc $ 

 $  = {a^3} + {b^3} + {c^3} + 3ab\left( {a + c} \right) + 3ac\left( {a + c} \right) + 3{b^2}\left( {a + c} \right) + 3bc\left( {a + c} \right) $ 

 $  = {a^3} + {b^3} + {c^3} + \left( {a + c} \right)\left( {3ab + 3ac + 3{b^2} + 3bc} \right) $ 

 $  = {a^3} + {b^3} + {c^3} + \left( {a + c} \right)\left( {3a\left( {b + c} \right) + 3b\left( {b + c} \right)} \right) $ 

 $  = {a^3} + {b^3} + {c^3} + \left( {a + c} \right)\left( {b + c} \right)\left( {3a + 3b} \right) $ 

 $  = {a^3} + {b^3} + {c^3} + 3\left( {a + c} \right)\left( {b + c} \right)\left( {a + b} \right) $ 

Now, on transposing we get

 $ {\left( {a + b + c} \right)^3} - {a^3} - {b^3} - {c^3} = 3\left( {a + b} \right)\left( {b + c} \right)\left( {c + a} \right) $ 

Hence, proved.