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

Answer

$\text { L.H.S }=x^3+y^3+z^3-3 x y z$
$=(x+y+z)\left(x^2+y^2+z^2-x y-y z-z x\right)$
(Using Identity $a^3+b^3+c^3-3 a b c=(a+b+c)\left(a^2+b^2+c^2-a b-b c-c a\right)$ )
$=\frac{1}{2}(x+y+z)\left\{2\left(x^2+y^2+z^2-x y-y z-z x\right)\right\} \text { (Multiplying and Dividing by 2) }$
$=\frac{1}{2}(x+y+z)\left(2 x^2+2 y^2+2 z^2-2 x y-2 y z-2 z x\right)$
$=\frac{1}{2}(x+y+z)\left\{\left(x^2-2 x y+y^2\right)+\left(y^2-2 y z+z^2\right)+\left(z^2-2 z x+x^2\right)\right\}$
$\left.=\frac{1}{2}(x+y+z)\left[(x-y)^2+(y-z)^2+(z-x)^2\right] . \text { (Using Identity }(a-b)^2=a^2-2 a b+b^2\right)$

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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 $x^3+3 x^2+3 x+1$ is divided by $x$.

Answer

Let $p(x)=x^3+3 x^2+3 x+1$
$x .$
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 $x^4+x^3-2 x^2+x+1$ is divided by $x-1$

Answer

Here, $p(x)=x^4+x^3-2 x^2+x+1$, 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 $x^4+x^3-2 x^2+x+1$ is divided by $x-1$

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Question 73 Marks

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

Answer

Let $p(x)=x^3-23 x^2+142 x-120$
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 $x^3-23 x^2+142 x-120=x^3-x^2-22 x^2+22 x+120 x-120$
$=x^2(x-1)-22 x(x-1)+120(x-1)$
$=(x-1)\left(x^2-22 x+120\right)[\text { Taking }(x-1) \text { common }]$
Now $x^2-22 x+120$ can be factorised either by splitting the middle term or by using the Factor theorem. By splitting the middle term, we have:
$x^2-22 x+120=x^2-12 x-10 x+120$
$=x(x-12)-10(x-12)$
$=(x-12)(x-10)$
Therefore, $x^3-23 x^2-142 x-120=(x-1)(x-10)(x-12)$

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Question 83 Marks

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

Answer

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

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Question 93 Marks

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

Answer

Given polynomial is $q(t)=4 t^3+4 t^2-t-1$
Let $g(t)=2 t+1$
For the zero of $g(t)$ put $g(t)=0$
$\therefore 2 t+1=0 \Rightarrow t=-1 / 2$
On putting $t=-\frac{1}{2}$ in $\mathrm{q}(\mathrm{t})$, 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, $(2 t+1)$ is a factor of $q(t)$, i.e. $q(t)$ is a multiple of $(2 t+1)$.

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