3 Marks Question

Question 13 Marks

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

Answer

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

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Question 23 Marks

Find the following product: $(2x - y + 3z) (4x^2 + y^2 + 9z^2 + 2xy + 3yz - 6xz)$

Answer

$(2x - y + 3z) (4x^2 + y^2 + 9z^2 + 2xy + 3yz - 6xz)$
$= {2x + (-y) + 3z}{(2x)^2 + (-y)^2 + (3z)^2 - (2x)(-y) - (-y)(3z) - (3z)(2x)}$
$= (2x)^3 + (-y)^3 + (3z)^3 - 3(2x)(-y)(3z)$
$\big[\therefore (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca) = a^3 + b^3 + c^3 - 3abc \big]$
$= 8x^3 - y^3 + 27z^2 + 18xyz$

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Question 33 Marks

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

Answer

Given $x - 2 = 0$
$ \Rightarrow x = 2$ Now, we have two equations
$3x^2 + 6x – 24$ 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.
$4x^2+ x – 2$ 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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MCQ 43 Marks

For the polynomial $\frac{\text{x}^3+2\text{x}+1}{5}-\frac{7}{2}\text{x}^2-\text{x}^6,$ write.

A
The degree of the polynomial.
B
The coefficient of $x^3.$
✓
The coefficient of $x^{6.}$
D
The constant term.

Answer

Correct option: C.

The coefficient of $x^{6.}$

$\frac{\text{x}^3+2\text{x}+1}{5}-\frac{7}{2}\text{x}^2-\text{x}^6$
$\frac{\text{x}^3}{5}+\frac{2\text{x}}{5}+\frac{1}{5}-\frac{7}{2}\text{x}^2-\text{x}^6$

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 $6$.
The coefficient of $x^3$ is $\frac{1}{5}$
The coefficient of $x^{6 }$is $-1$.
The constant term is $\frac{1}{5}$

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Question 53 Marks

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

Answer

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

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Question 63 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 73 Marks

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

Answer

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

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Question 83 Marks

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

Answer

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

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Question 93 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 103 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

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

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Question 113 Marks

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

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$, p - 1$ is a factor of $p^{10} - 1.$
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, $p - 1$ is a factor of $p^{11} - 1.$
Hence$, p - 1$ is a factor of$ p^{10} - 1$ and also of $p^{11} - 1.$

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Question 123 Marks

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

Answer

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

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Question 133 Marks

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

Answer

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

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Question 143 Marks

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

Answer

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

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Question 153 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 163 Marks

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

Answer

Let $p(x) = x^3 - 6x^2 + 11x - 6$
Constant term of $p(x) = -6$
$\therefore$ Factors of $-6$ are $±1, ±2, ±3, ±5, ±6$
By trial, we find that $p(1) = 0,$ so $(x - 1)$ is a factor of $p(x).$
$[\therefore (1)^3 - 6(1)^2 + 11(1) - 6 = 1 - 6 + 11 - 6 = 0]$
Now, we see that $x^3 - 6x^2 + 11x - 6$
$= x^3 - x^2 - 5x^2 + 5x + 6x - 6$
$= x^2 (x - 1) - 5x(x - 1) + (6x - 1)$
$= (x - 1)(x^2 - 5x + 6) [$taking$ (x - 1)$ common factor$]$
Now$, (x^2 - 5x + 6) = x^2 - 3x - 2x + 6 [$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 173 Marks

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

Answer

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

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Question 183 Marks

If both $x - 2$ and $\text{x}-\frac{1}{2}$ are factors of $px^2 + 5x + r,$ 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 193 Marks

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

Answer

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

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