To do:We have to find the remainder when $x^3+ 3x^2 + 3x + 1$ is divided by (i) \( x+1 \)(ii) \( x-\frac{1}{2} \)(iii) \( x \)(iv) \( x+\pi \)(v) \( 5+2 x \)Solution:The remainder theorem states that when a polynomial $p(x)$ is divided by a linear polynomial, $x - ... Read More

To do:We have to find the remainder when \( x^{3}-a x^{2}+6 x-a \) is divided by \( x-a \).Solution:The remainder theorem states that when a polynomial $p(x)$ is divided by a linear polynomial, $x - a$ the remainder of that division will be equivalent to $p(a)$.Let $f(x) =x^{3}-a x^{2}+6 x-a$ ... Read More

To do:We have to find whether polynomial $g(x)$ is a factor of polynomial $p(x)$ in each of the given cases.Solution:We know that, if $g(x)$ is a factor of $p(x)$, then the remainder will be zero.(i) \( p(x)=2 x^{3}+x^{2}-2 x-1, g(x)=x+1=x-(-1) \)So, the remainder will be $p(-1)$.$p(-1) = 2 (-1)^{3}+(-1)^{2}-2 (-1)-1$$= ... Read More

To do:We have to find the value of \( k \), if \( x-1 \) is a factor of \( p(x) \) in each of the given cases.Solution :Factor Theorem:The factor theorem states that if $p(x)$ is a polynomial of degree $n >$ or equal to $1$ and $a$ is ... Read More

To do:We have to factorise the given expressions.Solution:We can find the factorisation of the given expressions by using the splitting the middle term method.(i) \( 12 x^{2}-7 x+1 \)We have to find two numbers whose sum is $-7$ and the product is $1\times12 = 12$Here, $(-3)+(-4)=-7$ and $(-3)\times(-4)=12$.Therefore, We get ... Read More

To do:We have to factorise the given expressions.Solution:(i) \( x^{3}-2 x^{2}-x+2 \)Let $f(x)=x^{3}-2 x^{2}-x+2$Factors of $2$ are $\pm 1$ and $\pm 2$.     [Product of coefficient of $x^3$ and constant term $=1\times2=2$]By trial and error method, we get, $f(1)=(1)^{3}-2 (1)^{2}-(1)+2$$=1-2(1)-1+2$$=3-3$$=0$This implies, $x-1$ is a factor of $f(x)$.Therefore, $f(x)=x^3-2x^2-x+2$$= x^3 - ... Read More

To do:We have to use suitable identities to find the given products.Solution:We know that, $(p + a)(p+b) = p( p+ b) + a(p+b)$$=p^2+pb+ap+ab$$=p^2+p(a+b)+ab$Therefore, (i) \( (x+4)(x+10) \)Here, $p=x, a=4$ and $b=10$This implies, $(x+4)(x+10) = x^2+x(4+10)+4\times10$$=x^2+14x+40$Hence, $(x+4)(x+10)=x^2+14x+40$.(ii) \( (x+8)(x-10) \)Here, $p=x, a=8$ and $b=-10$This implies, $(x+8)(x-10) = x^2+x(8-10)+8\times(-10)$$=x^2-2x-80$Hence, $(x+8)(x-10)=x^2-2x-80$.(iii) \( (3 x+4)(3 x-5) ... Read More

To do:We have to evaluate the given products without multiplying directly.Solution:We know that, $(x + a)(x+b) = x( x+ b) + a(x+b)$$=x^2+xb+ax+ab$$=x^2+x(a+b)+ab$Therefore, (i) \( 103 \times 107=(100+3) \times (100+7) \)Here, $x=100, a=3$ and $b=7$This implies, $(100+3)(100+7) = (100)^2+100(3+7)+3\times7$$=10000+100(10)+21$$=10000+1000+21$$=11021$Hence, $103 \times 107=11021$.(ii) \( 95 \times 96=(100-5)\times(100-4) \)Here, $x=100, a=-5$ and $b=-4$This implies, ... Read More

To do:We have to factorise the given expressions using appropriate identities.Solution:(i) \( 9 x^{2}+6 x y+y^{2} \)\( 9 x^{2}+6 x y+y^{2} \) can be written as, $9 x^{2}+6 x y+y^{2}=(3x)^2+2\times(3x)\times(y)+(y)^2$Using the identity $a^2+2ab+b^2 = (a+b)^2$Here, $a = 3x$ and $b = y$Therefore, $9x^2+6xy+y^2 = (3x)^2+2(3x)(y)+y^2$$= (3x+y)^2$$= (3x+y)(3x+y)$(ii) \( 4 y^{2}-4 ... Read More

To do:We have to expand each of the given expressions using suitable identities.Solution:We know that, $(a+b+c)^2=a^2+b^2+c^2+2ab+2bc+2ca$Therefore, (i) \( (x+2 y+4 z)^{2} \)Here, $a=x, b=2y$ and $c=4z$$(x+2 y+4 z)^{2}=(x)^2+(2y)^2+(4z)^2+2(x)(2y)+2(2y)(4z)+2(4z)(x)$$=x^2+4y^2+16z^2+4xy+16yz+8xz$(ii) \( (2 x-y+z)^{2} \)Here, $a=2x, b=-y$ and $c=z$$(2 x-y+z)^{2}=(2x)^2+(-y)^2+(z)^2+2(2x)(-y)+2(-y)(z)+2(z)(2x)$$=4x^2+y^2+z^2-4xy-2yz+4xz$(iii) \( (-2 x+3 y+2 z)^{2} \)Here, $a=-2x, b=3y$ and $c=2z$$(-2 x+3 y+2 z)^{2}=(-2x)^2+(3y)^2+(2z)^2+2(-2x)(3y)+2(3y)(2z)+2(2z)(-2x)$$=4x^2+9y^2+4z^2-12xy+12yz-8xz$(iv) ... Read More