3 Marks Question

Question 13 Marks

Find the value of the polynomial $3 x^3-4 x^2+7 x-5$, when $x=3$ and also when $x=-3$

Answer

Let $\mathrm{p}(\mathrm{x})=3 \mathrm{x}^2-4 \mathrm{x}^2+7 \mathrm{x}-5 $
$\therefore \mathrm{p}(3)=3(3)^2-4(3)^2+7(3)-5=3(27)-4(9)+21-5=81-36+21-5=61$
Now, $\mathrm{p}(-3)$
$=3(-3)^2-4(-3)^2+7(-3)-5=3(-27)-4(9)-21-5=-81-36-21-5=-143$

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Question 23 Marks

Factorise:
$x^3+x^2-4 x-4$

Answer

Let $p(x)=x^3+x^2-4 x-4$
Constant term of $p(x)=-4$
$\therefore$ Factors of $-4$ are $\pm 1, \pm 2, \pm 4$
By trial, we find that $p(-1)=0$, so $(x+1)$ is a factor of $p(x)$.
Now, we see that $x^3+x^2-4 x-4$
$=x^2(x+1)-4(x+1)$
$=(x+1)\left(x^2-4\right)[\text { taking }(x+1) \text { common factor] }$
Now, $x^2-4=x^2-2^2$
$=(x+2)(x-2)\left[\text { Using identity, } a^2-b^2=(a-b)(a+b)\right]$
$\therefore x^3+x^2-4 x-4=(x+1)(x-2)(x+2)$

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Question 33 Marks

If $x+2 a$ is a factor of $x^5-4 a^2 x^3+2 x+2 a+3$, find $a$.

Answer

Let $p(x)=x^5-4 a^2 x^3+2 x+2 a+3$
If $x-(-2 a)$ is a factor of $p(x)$, then $p(-2 a)=0 $
$\therefore p(-2 a)=(-2 a)^5-4 a^2(-2 a)^3+2(-2 a)+2 a+3$
$=-32 a^5+32 a^5-4 a+2 a+3=-2 a+3$ Now, $p(-2 a)=0 $
$\Rightarrow-2 a+3=0 $
$\Rightarrow a=\frac{3}{2}$

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Question 43 Marks

Find the following product: $(2 x-y+3 z)\left(4 x^2+y^2+9 z^2+2 x y+3 y z-6 x z\right)$

Answer

$(2 x-y+3 z)\left(4 x^2+y^2+9 z^2+2 x y+3 y z-6 x z\right)$
$=\{2 x+(-y)+3 z\}(2 x)^2+(-y)^2+(3 z)^2-(2 x)(-y)-(-y)(3 z)-(3 z)(2 x)$
$=(2 x)^3+(-y)^3+(3 z)^3-3(2 x)(-y)(3 z)$
${\left[\therefore(a+b+c)\left(a^2+b^2+c^2-a b-b c-c a\right)=a^3+b^3+c^3-3 a b c\right]=8 x^3-y^3+27 z^2+18 x y z}$

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Question 53 Marks

Determine which of the following polynomials has $x - 2$ a factor:
$i. 3x^2 + 6x - 24$
$ii. 4x^2+ x – 2$

Answer

Given $x - 2 = 0$
$\Rightarrow x = 2$
Now, we have two equations
$i. 3x^2 + 6x – 24$
Put $x = 2$ in this equation
$3(2)^2 + 6 \times 2 - 24$
$3 \times 4 + 12 - 24$
$12 + 12 - 24 = 0$
$X - 2$ is a polynomial factor for this equation.
$ii. 4x^2+ x – 2$
Put $x = 2$ in this equation
$4(2)^2 + 2 - 2$
$4 \times 4 + 2 - 2 = 16$
$X - 2$ is a not a polynomial factor for this equation.
Hence, the equation $3x^2 + 6x – 24$ is an equation which has a polynomial factor of $x - 2.$

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Question 63 Marks

Factorise:
$3x^3 - x^2 - 3x + 1$

Answer

Let $p(x)=3 x^3-x^2-3 x+1$
Constant term of $p ( x )=-1$
$\therefore$ Factors of $1$ are $\pm 1$
By trial, we find that $p(1)=0$, so $(x-1)$ is a factor of $p(x)$.
Now, we see that $3 x^3-x^2-3 x-1$
$=3 x^2-3 x^2+2 x^2-2 x-x+1$
$=3 x^2(x-1)+2 x(x-1)-1(x-1)$
$=(x-1)\left(3 x^2+2 x-1\right)$
Now, $\left(3 x^2+2 x-1\right)=3 x^2-x^2-3 x-1$ [bi spliting middle term]
$=3 x^2(x+1)-1(x+1)=(x+1)(3 x-1)$
$\therefore 3 x^2-x^2-3 x-1=(x-1)(x+1)(3 x-1)$

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Question 73 Marks

Simplify $(2 x-5 y)^3-(2 x+5 y)^3$.

Answer

$\left.(2 x-5 y)^3-(2 x+5 y)^3=\{(2 x-5 y)-(2 x+5 y)\}(2 x-5 y)^2+(2 x-5 y)(2 x+5 y)+(2 x+5 y)^2\right\}$
${\left[\therefore a^3-b^3=(a-b)\left(a^2+a b+b^2\right)\right]}$
$=(2 x-5 y-2 x-5 y)\left(4 x^2+25 y^2-20 x y+4 x^2-25 y^2+4 x^2+25 y^2+20 x y\right)$
$=(-10 y)\left(2 x^2+25 y^2\right)$
$=-120 x^2 y-250 y^3$

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Question 83 Marks

For the polynomial $\frac{\text{x}^3+2\text{x}+1}{5}-\frac{7}{2}\text{x}^2-\text{x}^6,$ write.
$i.$ The degree of the polynomial.
$ii.$ The coefficient of $x^3.$
$iii.$ The coefficient of $x^{6.}$
$iv.$ The constant term.

Answer

$\frac{\text{x}^3+2\text{x}+1}{5}-\frac{7}{2}\text{x}^2-\text{x}^6$ $\frac{\text{x}^3}{5}+\frac{2\text{x}}{5}+\frac{1}{5}-\frac{7}{2}\text{x}^2-\text{x}^6$
$i.$ We know that highest power of variable in a polynomial is the degree of the polynomial. In the given polynomial, the term with highest of $x$ is $- x $, and the exponent of $x$ in this term in $6$ .
$ii.$ The coefficient of $x^3$ is $\frac{1}{5}$
$iii.$ The coefficient of $x^6$ is $-1 .$
$iv.$ The constant term is $\frac{1}{5}$

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Question 93 Marks

Check whether $p(x)$ is a multiple of $g(x)$ or not: $p(x)=2 x^3-11 x^2-4 x+5, g(x)=2 x+1$

Answer

$\mathrm{p}(\mathrm{x}) \text { will be a multiple } \mathrm{g}(\mathrm{x}) \text { if } \mathrm{g}(\mathrm{x}) \text { divides } \mathrm{p}(\mathrm{x}) \text {. Now, } \mathrm{g}(\mathrm{x})=2 \mathrm{x}+1 \text { give } \mathrm{x}=-\frac{1}{2} \text { Remainder }$
$=\mathrm{p}\left(-\frac{1}{2}\right)=2\left(\frac{-1}{2}\right)^3-11\left(\frac{-1}{2}\right)^2-4\left(\frac{-1}{2}\right)+5=2\left(\frac{-1}{8}\right)-11\left(\frac{1}{4}\right)^2+2+5=\frac{-1}{4}-\frac{11}{4}+7$
$=\frac{-1-11+28}{4}=\frac{16}{4}=4 \text { Since remainder } \neq 0 \text {, So } \mathrm{p}(\mathrm{x}) \text { is not a multiple of } \mathrm{g}(\mathrm{x})$.

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Question 103 Marks

If both $x -2$ and $x -\frac{1}{2}$ are factors of $px ^2+5 x + r$, show that $p = r$.

Answer

Let $f(x) = p(x)^2 + 5x + r$ Since, $x - 2$ is a factor of f(x), then $f(2) = 0$
$\therefore$ $p(2)^2 + 5(2) + r = 0$
$\Rightarrow 4p + 10 + r = 0 ...(i)$ Since, $\text{x}-\frac{1}{2}$ is a factor of
f(x), then $\text{f}\Big(\frac{1}{2}\Big)=0$
$\therefore\text{p}\Big(\frac{1}{2}\Big)^2+5\Big(\frac{1}{2}\Big)+\text{r}=0$
$\Rightarrow\text{p}\times\frac{1}{4}+\frac{5}{2}+\text{r}=0$
$\Rightarrow\text{p}+10+4\text{r}=0\ ...\text{(i)}$
Since, x - 2 and $\text{x}-\frac{1}{2}$ is a factors of
$f(x) = px^2+ 5x + r$. From Eqs. (i) and (ii),
$4p + 10 + r = p + 10 + 4r$
$\Rightarrow 3p = 3r$
$\therefore p = r$

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Question 113 Marks

Show that: $2x - 3$ is a factor of $x + 2x^3 - 9x^2 + 12.$

Answer

Let $p(x) = x + 2x^3 - 9x^2 + 12, g(x) = 2x - 3 g(x) = 2x - 3 = 0$
gives $\text{x}=\frac{3}{2}$ g(x) will be a factor of p(x) if $\text{p}\Big(\frac{3}{2}\Big)=0$ (Factor theorem)
Now, $\text{p}\Big(\frac{3}{2}\Big)=\frac{3}{2}+2\Big(\frac{3}{2}\Big)^3-9\Big(\frac{3}{2}\Big)^2+12$
$=\frac{3}{2}+2\Big(\frac{27}{8}\Big)-9\Big(\frac{9}{4}\Big)+12$
$=\frac{3}{2}+\frac{27}{4}-\frac{81}{4}+12$
$=\frac{6+27-81+48}{4}=\frac{0}{4}=0$
Since, $\text{p}\Big(\frac{3}{2}\Big)=0,$
So $g(x)$ is a factor of $p(x)$.

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Question 123 Marks

If $x+1$ is a factor of $a x^3+x^2-2 x+4 a-9$, find the value of $a$.

Answer

Let $p(x)=a x^3+x^2-2 x+4 a-9$ As $(x+1)$ is a factor of $p(x)$
$\therefore p(-1)=0\left[\right.$ By factor theorem]
$\Rightarrow a(-1)^3+(-1)^2-2(-1)+4 a-9=0 $
$\Rightarrow a(-1)+1+2+4 a-9=0 $
$\Rightarrow-a+4 a-6=0 $
$\Rightarrow 3 a-6=0 $
$\Rightarrow 3 a=6 $
$\Rightarrow a=2$

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Question 133 Marks

Factorise: $2x^3 - 3x^2 - 17x + 30$

Answer

Let $p ( x )=2 x ^3-3 x ^2-17 x +30$ Constant term of $p ( x )=30$
$\therefore$ Factors of $30$ are $\pm 1, \pm 2, \pm 3, \pm 5, \pm 6, \pm 10, \pm 15, \pm 30$ By trial,
we find that $p(2)=0$, so $(x-2)$ is a factor of $p(x)$.
$\left[\therefore 2(2)^3-3(2)^2-17(2)+30=16-12-34+30=0\right]$
Now, we see that $2 x^3-3 x^2-17 x+30=2 x^3-4 x^2+x^2-2 x-15 x+30$
$=2 x^2(x-2)+x(x-2)-15(x-2)$
$=(x-2)\left(2 x^2+x-15\right)[\text { taking }(x-2) \text { common factor }]$
Now, $\left(2 x^2+x-15\right)$ can be factorised either by spliting the middle term or by using the factor theorem.
Now, $\left(2 x^2+x-15\right)=2 x^2+6 x-5 x-15=2 x(x+3)-5(x+3)$ [by spliting the middle term]
$=(x+3)(2 x-5)$
$\therefore 2 x^3-3 x^2-17 x+30=(x-2)(x+3)(2 x-5)$

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Question 143 Marks

Find the value of m so that $2x - 1$ be a factor of $8x^4 + 4x^3 - 16x^2+ 10x + m.$

Answer

Let $p(x) = 8x^4 + 4x^3 - 16x^2+ 10x + m$
Since, $2x - 1$ is a factor of p(x), then put $\text{p}\Big(\frac{1}{2}\Big)=0$
$\therefore8\Big(\frac{1}{2}\Big)^4+4\Big(\frac{1}{2}\Big)^3-16\Big(\frac{1}{2}\Big)^2+10\Big(\frac{1}{2}\Big)+\text{m}=0$
$\Rightarrow8\times\frac{1}{16}+4\times\frac{1}{8}-16\times\frac{1}{4}+10\Big(\frac{1}{2}\Big)+\text{m}=0$
$\Rightarrow\frac{1}{2}+\frac{1}{2}-4+5+\text{m}=0$
$\Rightarrow1+1+\text{m}=0$
$\therefore\text{m}=-2$
Hence, the value of m is $-2.$

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Question 153 Marks

If the polynomials $az^3 + 4z^2 + 3^z - 4$ and $z^3 - 4z + a$ leave the same remainder when divided by $z - 3$, find the value of a.

Answer

$p(z)=a z^3+4 z^2+3^z-4 p(z)=z^3-4 z+a$
As these two polynomials leave the same remainder,
when divided by $z-3$, then $p(3)=q(3)$.
$\therefore p(3)=a(3)^3+4(3)^2+3(3)-4=27 a+36+9-4 \text { or } p(3)$
$=27 a+41 \text { And } q(3)=(3)^3-4(3)+a$
$=27-12+a=15+1$
Now, $p(3)=q(3)$
$\Rightarrow 27 a+41=15+a$
$\Rightarrow 26 a=-26 a, a=-1$
Hence, the required value of $a=-1$.

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Question 163 Marks

If $a + b + c = 9$ and $ab + bc + ca = 26$, find $a^2 + b^2 + c^2.$

Answer

We have that $(a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + bc + 2ca$
$\Rightarrow (a + b + c)^2 = (a^2 + b^2 + c^2) + 2(ab + bc + ca)$
$\Rightarrow 9^2 = (a^2 + b^2 + c^2) + 2(26)$ [Putting the value of $a + b + c$ and $ab + bc + ca]$
$\Rightarrow 81 = (a^2 + b^2 + c^2) + 52$
$\Rightarrow (a^2 + b^2 + c^2) = 81 - 52 = 29$

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Question 173 Marks

Show that $\mathrm{p}-1$ is a factor of $\mathrm{p}^{10}-1$ and also of $\mathrm{p}^{11}-1$.

Answer

If $p-1$ is a factor of $p^{10}-1$, then $(1)^{10}-1$ should be equal to zero.
Now, $(1)^{10}-1=1-1=0$
Therefore, $\mathrm{p}-1$ is a factor of $\mathrm{p}^{10}-1$.
Again, if $p-1$ is a factor of $p^{11}-1$, then $(1)^{11}-1$ should be equal to zero.
Now, $(1)^{11}-1=1-1=0$
Therefore, $\mathrm{p}-1$ is a factor of $\mathrm{p}^{11}-1$.
Hence, $p-1$ is a factor of $p^{10}-1$ and also of $p^{11}-1$.

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Question 183 Marks

Factorise: $x^3 - 6x^2 + 11x - 6$

Answer

Let $p(x)=x^3-6 x^2+11 x-6$ Constant term of $p(x)=-6$
$\therefore$ Factors of $-6$ are $\pm 1, \pm 2, \pm 3, \pm 5, \pm 6$ By trial,
we find that $p(1)=0$, so $(x-1)$ is a factor of $p(x)$.
$\left[\therefore(1)^3-6(1)^2+11(1)-6=1-6+11-6=0\right]$
Now, we see that $x^3-6 x^2+11 x-6=x^3-x^2-5 x^2+5 x+6 x-6$
$=x^2(x-1)-5 x(x-1)+(6 x-1)=(x-1)\left(x^2-5 x+6\right)[$ taking $(x-1)$ common factor $]$
Now, $\left(x^2-5 x+6\right)=x^2-3 x-2 x+6[$ by spliting middle term]
$= x(x - 3) -2(x - 3) = (x - 3)(x - 2) $
$\therefore x^3 - 6x^2 + 11x - 6 = (x - 1)(x - 2)(x - 3)$

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Question 193 Marks

If $a + b + c =5$ and $ab + bc + ca =10$, then prove that $a ^3+ b ^3+ c ^3-3 abc =-25$.

Answer

To prove, $a^3+b^3+c^3-3 a b c=-25$
Given, $a+b+c=5, a b+b c+c a=10$
$\therefore(a+b+c)^2=a^2+b^2+c^2+2(a b+b c+c a)$
$\therefore$ $(5)^2 = a^2 + b^2 + c^2 = 25(10)$
$\Rightarrow 25 = a^2 + b^2 + c^2 = 20$
$\Rightarrow a^2 + b^2 + c^2 = 25 - 20$
$\Rightarrow a^2 + b^2 + c^2 = 5$
$LHS = a^3 + b^3+ c^3 - 3abc$
$= (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca)$
$= (5)[5 - (ab + bc + ca)]$
$= 5(5 - 10) = 5(-5) = -25$
$= RHS$ Hence proved.

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