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#2.4

For which values of a and b, are the zeroes of $q(x) = x^3 + 2x^2 + a$ also the zeroes of the polynomial $p(x) = x^5 - x^4 - 4x^3 + 3x^2 + 3x + b$ ? Which zeroes of p(x) are not the zeroes of q(x)?

Answer. [2, 1]

Solution.

Given :-

$q(x) = x^3 + 2x^2 + a$ , $p(x) = x^5 - x^4 - 4x^3 + 3x^2 + 3x + b$

Hence, q(x) is a factor of p(x) use divided algorithm

Since $x^3 + 2x^2 + a$ is a factor hence remainder = 0

$x^2(-1 - a) + 3x(1 + a) + (b - 2a) = 0.x^2 + 0.x + 0$

By comparing

– 1 – a = 0 b – 2a = 0

a = – 1 2a = b

put a = –1

2(–1) = b

b = –2

Hence, a = –1, b = –2

$x^5 - x^4 - 4x^3 + 3x^2 + 3x - 2 = (x^3 + 2x^2 - 1) (x^2 - 3x + 2)$

$= (x^3 + 2x^2- 1) (x^2 - 2x - x + 2)$

$= (x^3 + 2x^2 - 1) (x(x - 2) - 1(x - 2))$

$= (x^3 + 2x^2 -1) (x - 2) (x - 1)$

x – 2 = 0 x – 1 = 0

x = 2 x = 1

Hence, 2, 1 are the zeroes of p(x) which are not the zeroes of q(x).

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#2.4

Given that $x-\sqrt{5}$ is a factor of the cubic polynomial $x^3 -3 \sqrt{5}x^2 + 13x - 3\sqrt{5}$ ,find all the zeroes of the polynomial.

Answer. $\sqrt{5},\sqrt{5}+\sqrt{2}, \sqrt{5}-\sqrt{2}$

Solution.

The given cubic polynomial is $x^3 -3 \sqrt{5}x^2 + 13x - 3\sqrt{5}$

Hence $x-\sqrt{5}$ is a factor

Use divided algorithm

$x^3 - 3\sqrt{3} x^2 + 13x - 3\sqrt{5}=(x-\sqrt{5}) (x^{2}-2\sqrt{5}+3)$

$=(x-\sqrt{5}) (x^{2}-2\sqrt{5}+3)$

$x^{2}-2\sqrt{5}+3$

a = 1, b = $-2\sqrt{5}$ , c = 3

$x=\frac{-b\pm \sqrt{b^{2}}-4ac}{2a}$

$x=\frac{2\sqrt{5}\pm \sqrt{20-12}}{2}$

$x=\frac{2\sqrt{5}\pm 2\sqrt{2}}{2}=\sqrt{5}\pm \sqrt{2}$

$x=\sqrt{5}+\sqrt{2}, \sqrt{5}-\sqrt{2}$

Hence, $\sqrt{5},\sqrt{5}+\sqrt{2}, \sqrt{5}-\sqrt{2}$ are the zeroes of the polynomial.

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#2.4

Find k so that $x^2 + 2x + k$ is a factor of $2x^4 + x^3 - 14x^2 + 5x + 6$ . Also find all the zeroes of the two polynomials.

The given polynomial is $2x^4 + x^3 - 14x^2 + 5x + 6$

Here, $x^2 + 2x + k$ is a factor of $2x^4 + x^3 - 14x^2 + 5x + 6$

Use division algorithm

Here, $x(7k + 21) + (6 + 8k + 2k^2)$ is remainder

It is given that $x^2 + 2x + k$ is a factor hence remainder = 0

$x(7k + 21) + (6 + 8k + 2k^2) = 0.x + 0$

By comparing L.H.S. and R.H.S.

7k + 21 = 0

k = – 3 ….(1)

$2k^2 + 8k + 6 = 0$

$2k^2 + 2k + 6k + 6 = 0$

2k(k + 1) + 6(k + 1) = 0

(k + 1) (2k + 6) = 0

k = –1, –3 …..(2)

From (1) and (2)

k = –3

Hence, $2x^4 + x^3 - 14x^2 + 5x + 6 = (x^2 + 2x + k) (2x^2 - 3x - 8 - 2k)$

Put k = –3

$= (x^2 + 2x - 3) (2x^2 - 3x - 8 + 6)$

$= (x^2 + 2x - 3) (2x^2 - 3x - 2)$

$= (x^2 + 3x - x - 3) (2x^2 - 4x + x - 2)$

$= (x(x + 3) - 1(x + 3)) (2x(x - 2) + 1(x - 2))$

$= (x + 3)(x -1)(x - 2) (2x + 1)$

x + 3 = 0 x – 1 = 0 x – 2 = 0 2x + 1 = 0

x = –3 x = 1 x = 2 2x = – 1

$x=\frac{-1}{2}$

Here zeroes of $x^2 + 2x - 3$ is –3, 1

Zeroes of $2x^2 - 3x - 2 is 2, \frac{-1}{2} .$

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#2.4

Given that $\sqrt{2}$ is a zero of the cubic polynomial $6x^3 + \sqrt{2}x^2 - 10x - 4\sqrt{2}$ , find its other two zeroes.

Answer. $\left [\frac{-2\sqrt{2}}{3}, \frac{-1}{\sqrt{2}} \right ]$

Given cubic polynomial is $6x^3 + \sqrt{2}x^2 - 10x - 4\sqrt{2}$

If $\sqrt{2}$ is a zero of the polynomial then (x – $\sqrt{2}$ ) is a factor of $6x^3 + \sqrt{2}x^2 - 10x - 4\sqrt{2}$

$6x^3 + \sqrt{2} x^2 - 10x -4\sqrt{2} = (6x^2 + 7\sqrt{2} x + 4) (x -\sqrt{2} )$

$= (6x^2 + 4\sqrt{2} x + 3\sqrt{2} x + 4\sqrt{2}) (x -\sqrt{2} )$

$= (2x(3x + 2\sqrt{2} ) +\sqrt{2} (3x + 2\sqrt{2} )) (x -\sqrt{2} )$

$= (3x + 2 \sqrt{2}) (2x +\sqrt{2} ) (x -\sqrt{2} )$ =0

$3x + 2\sqrt{2} = 0$ $2x + \sqrt{2} = 0$

$x=\frac{-2\sqrt{2}}{3}$ $x=\frac{-1}{\sqrt{2}}$

Hence, $\left [\frac{-2\sqrt{2}}{3}, \frac{-1}{\sqrt{2}} \right ]$ are the other two zeroes.

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#2.4

Given that the zeroes of the cubic polynomial $x^3 -6x^2 + 3x + 10$ are of the form a, a + b, a + 2b for some real numbers a and b, find the values of a and b as well as the zeroes of the given polynomial.

Answer. [5, 2, –1]

Solution. Here the given cubic polynomial is $x^3 -6x^2 + 3x + 10$

A = 1, B = –6, C = 3, D = 10

Given that, $\alpha = a, \beta = a + b, \gamma = a + 2b$

We know that, $\alpha +\beta +\gamma =\frac{-B }{A}$

$a + a + b + a + 2b =\frac{-(-6)}{1}$

$3a + 3b = 6$

3(a + b) = 6

a + b = 2 …..(1)

$\alpha \beta +\beta \gamma +\alpha = \frac{C}{A}$

$a(a + b) + (a + b) (a + 2b) + (a + 2b) (a) =\frac{3}{1}$

$a(2) + 2(a + 2b) + a^2 + 2ab = 3$ (Q a + b = 2)

$2a + 2a + 4b + a^2 + 2ab = 3$

$4a + 4b + a^2 + 2ab = 3$

$4a + 4(2 - a) + a^2 + 2a(2 - a) = 3$ (Q a + b = 2)

$4a + 8 - 4a + a^2 + 4a - 2a^2 = 3$

$a^2 + 4a - 2a^2 + 8 - 3 = 0$

$-a^2 + 4a + 5 = 0$

$a^2 - 4a - 5 = 0$

$a^2 - 5a + a - 5 = 0$

$a(a - 5)+1 (a - 5) = 0$

$(a -5) ( a + 1) = 0$

a = 5, –1

Put a = 5 in (1) put a = –1 in (1)

5 + b = 2 –1 + b = 2

b = –3 b = 3

Hence, value of a = 5, b = –3 and a = –1, b = 3

put a = 5, b = –3 and we get zeroes

a = 5

a + b = 5 – 3 = 2

a + 2b = 5 + 2(–3) = –1

Hence, zeroes are 5, 2, –1.

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#2.4

For each of the following, find a quadratic polynomial whose sum and product respectively of the zeroes are as given. Also find the zeroes of these polynomials by factorisation.

(i) $\frac{-8}{3},\frac{4}{3}$

(ii) $\frac{21}{8},\frac{5}{16}$

(iii) $-2\sqrt{3}, -9$

(iv) $\frac{-3}{2\sqrt{5}},-\frac{1}{2}$

(i) Answer. $\left [ -2 , -\frac{2}{3}\right ]$

Solution. Zeroes : zeroes of the polynomial are the value(s) that makes it equal to 0.

Here, sum of zeroes $=\frac{-8}{3}$

Product of zeroes $=\frac{4}{3}$

Let p(x) is the required polynomial

p(x) = x² – (sum of zeroes)x + (product of zeroes)

$=x^{2}-\left (\frac{-8}{3} \right )x+\frac{4}{3}$

$=x^{2}+\frac{8}{3}x+\frac{4}{3}$

Multiply by 3

$p(x) = 3x^2 + 8x + 4$

Hence, $3x^2 + 8x + 4$ is the required polynomial

$p(x) = 3x^2 + 8x + 4$

$= 3x^2 + 6x + 2x + 4$

$= 3x(x + 2) + 2(x + 2)$

$= (x + 2) (3x + 2)$ =0

$x = -2,\frac{2}{3}$ are the zeroes of p(x).

(ii) Answer. $\left [\frac{5}{2}, \frac{1}{8} \right ]$

Solution. Zeroes : zeroes of the polynomial are the value(s) that makes it equal to 0.

Here, sum of zeroes $=\frac{21}{8}$

Product of zeroes $=\frac{5}{16}$

Let p(x) is the required polynomial

p(x) = x² – (sum of zeroes)x + (product of zeroes)

$=x^{2}-\left (\frac{21}{8} \right )x+\frac{5}{16}$

$=x^{2}-\frac{21}{8} x+\frac{5}{16}$

Multiply by 16 we get

$p(x) = 16x^2 - 42x + 5$

Hence, $16x^2 - 42x + 5$ is the required polynomial

$p(x) = 16x^2 - 42x + 5$

$= 16x^2 - 40x -2x + 5$

= $8x(2x - 5) - 1(2x - 5)$

= $(2x - 5) (8x - 1)$ =0

$x=\frac{5}{2}, \frac{1}{8}$ are the zeroes of p(x).

(iii) Answer. $\left [-3\sqrt{3}, \sqrt{3} \right ]$

Solution. Zeroes : zeroes of the polynomial are the value(s) that makes it equal to 0.

Here, sum of zeroes $-2\sqrt{3}$

Product of zeroes = –9

Let p(x) is required polynomial

p(x) = x² – (sum of zeroes)x + (product of zeroes)

$=x^{2 }-(-2\sqrt{3})x+(-9)$

$=x^{2 } +2\sqrt{3}x+(-9)$

$p(x)=x^{2 } +2\sqrt{3}x+(-9)$

Hence, $x^{2 } +2\sqrt{3}x+(-9)$ is the required polynomial

$p(x)=x^{2 } +2\sqrt{3}x+(-9)$

= $x^{2 } +3\sqrt{3}x-\sqrt{3}x+(-9)$

= $x (x+3\sqrt{3})- \sqrt{3}(x + 3 \sqrt{3})$

= $(x+3\sqrt{3})(x - \sqrt{3})$ =0

$x =-3\sqrt{3}, \sqrt{3}$ are the zeroes of p(x)

(iv) Answer. $\left [\frac{\sqrt{5}}{2}, \frac{-1}{\sqrt{5}} \right ]$

Solution. Zeroes : zeroes of the polynomial are the value(s) that makes it equal to 0.

Here sum of zeroes $\frac{-3}{2\sqrt{5}}$

Product of zeroes $=-\frac{1}{2}$

Let p(x) is the required polynomial

p(x) = x² – (sum of zeroes)x + (product of zeros)

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

$p(x)=x^{2}+\frac{3}{2\sqrt{5}} x- \frac{1}{2}$

Multiplying by $2 \sqrt{5}$ we get

$p(x)=2 \sqrt{5}x^{2}+3x-\sqrt{5}$

Hence, $2 \sqrt{5}x^{2}+3x-\sqrt{5}$ is the required polynomial

$p(x)=2 \sqrt{5}x^{2}+3x-\sqrt{5}$

$=2 \sqrt{5}x^{2}+5x-2x -\sqrt{5}$

= $\sqrt{5}(2x+\sqrt{5})-1(2x+\sqrt{5})$

= $(\sqrt{5}x -1)(2x+\sqrt{5})$

$\left [\frac{-\sqrt{5}}{2}, \frac{1}{\sqrt{5}} \right ]$ are the zeroes of p(x).

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#2.4

prove that: $(a+b+c)^{3}-a^{3}-b^{3}-c^{3}=3(a+b)(b+c)(c+a)$

Taking left hand side

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

Now using the identity

$x^{3}-y^{3}=(x-y)(x^{2}+y^{2}+xy)$

So,

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

Now using $x^{3}-y^{3}=(x-y)(x^{2}+y^{2}+xy)$

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

So equation (i) becomes:

$\\(a+b+c)^{3}-a^{3}-b^{3}-c^{3}\\ =(a+b+c-a)\left [ (a+b+c)^{2}+a^{2}+a(a+b+c) \right ]-\left [ (b+c)(b^{2}+c^{2}-bc) \right ]\\ =(b+c)\left [ (a+b+c)^{2}+a^{2}+a(a+b+c)\right ]\left [ (b+c)(b^{2}+c^{2}-bc) \right ]\cdots \cdots \cdots (ii)\\$

Now,

$(a+b+c)^{3}-a^{3}-b^{3}-c^{3}\\ =(b+c)\left [ a^{2}+b^{2}+c^{2}+2ab+2bc+2ca+a^{2}+a^{2}+ab+ac-(b^{2}+c^{2}-bc) \right ]\\ =(b+c)\left [ b^{2}+c^{2}+3a^{2}+3ab+3ac-b^{2}-c^{2}+3bc \right ]\\ =(b+c)\left [ 3(a^{2})+ab+ac+bc \right ]\\ =3(b+c)\left [ a(a+b)+c(a+b) \right ]\\ =3(b+c)\left [ (a+b)(b+c) \right ]\\ =3(a+b)(b+c)(c+a)=R.H.S$

Hence proved.

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#2.4

If a + b + c = 5 and ab + bc + ca = 10, then prove that a³ + b³ + c³ –3abc = – 25

Given: (a + b + c) = 5, ab + bc + ca = 10

To Prove: a³ + b³ + c³ –3abc = – 25

We know that

$(a+b+c)^{2}=a^{2}+b^{2}+c^{2}+2(ab+bc+ca)\\ (5)^{2}=a^{2}+b^{2}+c^{2}+2(10)\\ 25=a^{2}+b^{2}+c^{2}+20\\ 25-20=a^{2}+b^{2}+c^{2}\\ 5=a^{2}+b^{2}+c^{2}$

Now,

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

Putting the values, we get:

$a^{3}+b^{3}+c^{3}-3abc=5\left [ 5-(ab+bc+ca) \right ]\\ =5(5-10)\\ =5(-5)\\ =-25\\=R.H.S$

Hence proved.

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#2.4

If a, b, c are all non-zero and a + b + c = 0, prove that $\frac{a^{2}}{bc}+\frac{b^{2}}{ca}+\frac{c^{2}}{ab}=3$

Given, $\frac{a^{2}}{bc}+\frac{b^{2}}{ca}+\frac{c^{2}}{ab}=3$

L.H.S. $\frac{a^{2}}{bc}+\frac{b^{2}}{ca}+\frac{c^{2}}{ab}$

Taking LCM of denominators, we get

L.H.S $\frac{a^{3}+b^{3}+c^{3}}{abc}$

Now we know that if a + b + c = 0 then $a^{3}+b^{3}+c^{3}=3abc$

Putting the value in above equation:

L.H.S = $\frac{3abc}{abc}$ = 3 = RHS

Hence proved.

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#2.4

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

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

Solution

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

This can be written as:

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

We know that

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

So comparing the RHS of equation (i) with the above identity:

a = x

b = -2y

c = -z

We get:

$\left ( x+(-2y)+(-z) \right )\left ((x)^{2}+(-2y)^{2}+(z)^{2}-(x)(-z)-(-z)(-2y)\right )\\ =(x)^{3}+(-2y)^{3}+(-z)^{3}-3(x)(-2y)(-z)\\ =x^{3}-8y^{3}-z^{3}-6xyz$

Hence the answer is $x^{3}-8y^{3}-z^{3}-6xyz$ .

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JEE Main high-scoring chapters and topics

Given that the zeroes of the cubic polynomial are of the form a, a + b, a + 2b for some real numbers a and b, find the values of a and b as well as the zeroes of the given polynomial.

For which values of a and b, are the zeroes of $q(x) = x^3 + 2x^2 + a$ also the zeroes of the polynomial $p(x) = x^5 - x^4 - 4x^3 + 3x^2 + 3x + b$ ? Which zeroes of p(x) are not the zeroes of q(x)?

Given that $x-\sqrt{5}$ is a factor of the cubic polynomial $x^3 -3 \sqrt{5}x^2 + 13x - 3\sqrt{5}$ ,find all the zeroes of the polynomial.

Find k so that $x^2 + 2x + k$ is a factor of $2x^4 + x^3 - 14x^2 + 5x + 6$ . Also find all the zeroes of the two polynomials.

Given that $\sqrt{2}$ is a zero of the cubic polynomial $6x^3 + \sqrt{2}x^2 - 10x - 4\sqrt{2}$ , find its other two zeroes.

Given that the zeroes of the cubic polynomial $x^3 -6x^2 + 3x + 10$ are of the form a, a + b, a + 2b for some real numbers a and b, find the values of a and b as well as the zeroes of the given polynomial.

prove that: $(a+b+c)^{3}-a^{3}-b^{3}-c^{3}=3(a+b)(b+c)(c+a)$

If a + b + c = 5 and ab + bc + ca = 10, then prove that a³ + b³ + c³ –3abc = – 25

If a, b, c are all non-zero and a + b + c = 0, prove that $\frac{a^{2}}{bc}+\frac{b^{2}}{ca}+\frac{c^{2}}{ab}=3$

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