3 Marks Question

Question 13 Marks

Write the cube in expanded form: ${\left( {2x + 1} \right)^3}$

Answer

${\left( {2x + 1} \right)^3}$
We know that ${\left( {a + b} \right)^3} = {a^3} + {b^3} + 3ab\left( {a + b} \right)$
$\therefore {\left( {2x + 1} \right)^3} = {\left( {2x} \right)^3} + {\left( 1 \right)^3} + 3 \times 2x \times 1\left( {2x + 1} \right)$
$ = 8{x^3} + 1 + 6x\left( {2x + 1} \right)\,$
$= 8{x^3} + 12{x^2} + 6x + 1.$
Therefore, the expansion of the expression ${\left( {2x + 1} \right)^3}$ is $8{x^3} + 12{x^2} + 6x + 1$

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Question 23 Marks

Verify that $x^3 + y^3 + z^3 - 3xyz = \frac{1}{2}(x + y + z)[(x - y)^2 + (y- z)^2 + (z - x)^2].$

Answer

$L.H.S = x^3 + y^3 + z^3 - 3xyz$
$= (x + y + z)(x^2 + y^2 + z^2 - xy - yz - zx)$
$($Using Identity $a^3+ b^3+ c^3– 3abc= (a + b + c)(a^2+ b^2+ c^2– ab – bc – ca))$
$= \frac{1}{2}(x + y + z){2(x^2 + y^2 + z^2 - xy - yz - zx)} ($Multiplying and Dividing by $2)$
$= \frac{1}{2}(x + y + z)(2x^2 + 2y^2 + 2z^2 - 2xy - 2yz - 2zx)$
$= \frac{1}{2}(x + y + z){(x^2 - 2xy + y^2) + (y^2 - 2yz + z^2)+ (z^2 - 2zx + x^2)}$
$= \frac{1}{2}(x + y + z)[(x - y)^2 + (y - z)^2 + (z - x)^2]. ($ Using Identity $(a – b)^2= a^2– 2ab + b^2)$

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Question 33 Marks

Use suitable identity to find the product:
$\left( {x + 4} \right)\left( {x + 10} \right)$

Answer

$\left( {x + 4} \right)\left( {x + 10} \right)$
We know that $\left( {x + a} \right)\left( {x + b} \right) = {x^2} + \left( {a + b} \right)x + ab$

Here a = 4 and b = 10

We need to apply the above identity to find the product $\left( {x + 4} \right)\left( {x + 10} \right) = {x^2} + \left( {4 + 10} \right)x + \left( {4 \times 10} \right)$
$= {x^2} + 14x + 40.$
Therefore, we conclude that the product$ \left( {x + 4} \right)\left( {x + 10} \right)$ is ${x^2} + 14x + 40$

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Question 43 Marks

Find the remainder when $x^3 + 3x^2 + 3x + 1$ is divided by $x$.

Answer

Let $p(x) = x^3 + 3x^2 + 3x + 1x$.
Remainder $= (0)^3+ 3(0)^2+ 3(0) + 1$
$ = 1$

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Question 53 Marks

Verify $x = - \frac{1}{{\sqrt 3 }},\frac{2}{{\sqrt 3 }}$ are zeroes of the polynomial $p\left( x \right) = 3{x^2} - 1$

Answer

$p\left( x \right) = 3{x^2} - 1,\,\,\,\,x = - \frac{1}{{\sqrt 3 }},\frac{2}{{\sqrt 3 }}$
We need to check whether $p\left( x \right) = 3{x^2} - 1{\text{ at }}x = - \frac{1}{{\sqrt 3 }},\frac{2}{{\sqrt 3 }}$ is equal to zero or not, i.e., $p\left( {{{ - 1} \over {\sqrt 3 }}} \right)$ and $p\left( {{2 \over {\sqrt 3 }}} \right)$ is equal to zero or not.
At $x = \frac{{ - 1}}{{\sqrt 3 }}$
$p\left( { - \frac{1}{{\sqrt 3 }}} \right) = 3{\left( { - \frac{1}{{\sqrt 3 }}} \right)^2} - 1\,\, = 3\left( {\frac{1}{3}} \right) - 1\,\, = 1 - 1\,\, = 0$
At $x = \frac{2}{{\sqrt 3 }}$
$p\left( {\frac{2}{{\sqrt 3 }}} \right) = 3{\left( {\frac{2}{{\sqrt 3 }}} \right)^2} - 1\,\, = 3\left( {\frac{4}{3}} \right) - 1\, = 4 - 1\, = 3$
Therefore, we can conclude that $x = \frac{{ - 1}}{{\sqrt 3 }}$ is a zero of the polynomial $p\left( x \right) = 3{x^2} - 1$ but $x = \frac{{ - 1}}{{\sqrt 3 }}$ is not a zero of the polynomial $p\left( x \right) = 3{x^2} - 1$

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Question 63 Marks

Find the remainder when $x^4 + x^3 – 2x^2 + x + 1$ is divided by $x – 1$

Answer

Here, $p(x) = x^4 + x^3 – 2x^2 + x + 1,$ and the zero of $x – 1$ is $1.$
Therefore, $p(1) = (1)^4 + (1)^3 – 2(1)^2 + 1 + 1 = 2$
Therefore, by the Remainder Theorem, $2$ is the remainder when $x^4 + x^3 – 2x^2 + x + 1$ is divided by $x – 1$

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Question 73 Marks

Factorise : $x^3 – 23x^2 + 142x – 120$

Answer

Let $p(x) = x^3 – 23x^2 + 142x – 120$
We shall now look for all the factors of $–120$. Some of these are $\pm 1, \pm 2, \pm 3, \pm 4, \pm 5, \pm 6, \pm 8, \pm 10, \pm 12, \pm 15, \pm 20, \pm 24, \pm 30, \pm 60.$
By hit and trial, we find that $p(1) = 0.$ Therefore, $x – 1$ is a factor of $ p(x).$
Now we see that $x^3 – 23x^2 + 142x – 120 = x^3 – x^2 – 22x^2 + 22x + 120x – 120$
$= x^2 (x –1) – 22x(x – 1) + 120(x – 1)$
$= (x – 1) (x^2 – 22x + 120) [$Taking $(x – 1)$ common$]$
Now $x^2 – 22x + 120$ can be factorised either by splitting the middle term or by using the Factor theorem. By splitting the middle term, we have: $x^2 – 22x + 120 = x^2 – 12x – 10x + 120$
$= x(x – 12) – 10(x – 12)$
$= (x – 12) (x – 10)$
Therefore, $x^3 – 23x^2 – 142x – 120 = (x – 1)(x – 10)(x – 12)$

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Question 83 Marks

Examine whether $x + 2$ is a factor of $x^3 + 3x^2 + 5x + 6$ and of $2x + 4.$

Answer

The zero of $x + 2$ is $–2.$
Let $p(x) = x^3 + 3x^2 + 5x + 6$ and $s(x) = 2x + 4$
Then, $p(–2) = (–2)^3 + 3(–2)^2 + 5(–2) + 6$
$= –8 + 12 – 10 + 6$
$= 0$
So, by the Factor Theorem, $x + 2$ is a factor of $x^3 + 3x^2 + 5x + 6.$
Again, $s(–2) = 2(–2) + 4 = 0$
So, $x + 2$ is a factor of $2x + 4.$

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Question 93 Marks

Check whether the polynomial $q(t) = 4t^3 + 4t^2 - t - 1$ is a multiple of $2t + 1$ or not.

Answer

Given polynomial is $q(t) = 4t^3 + 4t^2 - t - 1$
Let $g(t) = 2t + 1$
For the zero of $g(t)$ put $g(t) = 0$
$\therefore 2t + 1 = 0 $
$\Rightarrow t = -1/2$
On putting $t=-\frac{1}{2}$ in $q(t),$ we get
$q\left(\frac{-1}{2}\right)=4\left(\frac{-1}{2}\right)^{3}+4\left(\frac{-1}{2}\right)^{2}-\left(\frac{-1}{2}\right)-1$
$=4\left(\frac{-1}{8}\right)+4\left(\frac{1}{4}\right)+\frac{1}{2}-1$
$=-\frac{1}{2}+1+\frac{1}{2}-1 = 0$
At $t=-\frac{1}{2},$ we get $q\left(-\frac{1}{2}\right)=0$
i.e. the remainder obtained on dividing $q(t)$ by $g(t)$ is $0$.
Hence$, (2t +1)$ is a factor of $q(t), i.e. q(t)$ is a multiple of $(2t + 1).$

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