If $x+y+z=0$, show that $x^{3}+y^{3}+z^{3}=3 x y z$.
Solution: It is known that, $x^{3}+y^{3}+z^{3}-3 x y z=(x+y+z)\left(x^{2}+y^{2}+z^{2}-x y-y z-z x\right)$ Put $x+y+z=0$, $x^{3}+y^{3}+z^{3}-3 x y z=(0)\left(x^{2}+y^{2}+z^{2}-x y-y z-z x\right)$ $x^{3}+y^{3}+z^{3}-3 x y z=0$ $x^{3}+y^{3}+z^{3}=3 x y z$...
Read More →Verify that $x^{3}+y^{3}+z^{3}-3 x y z=\frac{1}{2}(x+y+z)\left[(x-y)^{2}+(y-z)^{2}+(z-x)^{2}\right]$
Solution: It is known that, $x^{3}+y^{3}+z^{3}-3 x y z=(x+y+z)\left(x^{2}+y^{2}+z^{2}-x y-y z-z x\right)$ $=\frac{1}{2}(x+y+z)\left[2 x^{2}+2 y^{2}+2 z^{2}-2 x y-2 y z-2 z x\right]$ $=\frac{1}{2}(x+y+z)\left[\left(x^{2}+y^{2}-2 x y\right)+\left(y^{2}+z^{2}-2 y z\right)+\left(x^{2}+z^{2}-2 z x\right)\right]$ $=\frac{1}{2}(x+y+z)\left[(x-y)^{2}+(y-z)^{2}+(z-x)^{2}\right]$...
Read More →Factorise: $27 x^{3}+y^{3}+z^{3}-9 x y z$
Solution: It is known that, $x^{3}+y^{3}+z^{3}-3 x y z=(x+y+z)\left(x^{2}+y^{2}+z^{2}-x y-y z-z x\right)$ $\therefore 27 x^{3}+y^{3}+z^{3}-9 x y z$ $=(3 x)^{3}+(y)^{3}+(z)^{3}-3(3 x)(y)(z)$ $=(3 x+y+z)\left[(3 x)^{2}+y^{2}+z^{2}-(3 x)(y)-(y)(z)-z(3 x)\right]$ $=(3 x+y+z)\left[9 x^{2}+y^{2}+z^{2}-3 x y-y z-3 x z\right]$...
Read More →Factorise each of the following: <br/><br/>(i) $27 y^{3}+125 z^{3}$<br/><br/>(ii) $64 m^{3}-343 n^{3}$
Solution: (i) $27 y^{3}+125 z^{3}$ $=(3 y)^{3}+(5 z)^{3}$ $=(3 y+5 z)\left[(3 y)^{2}+(5 z)^{2}-(3 y)(5 z)\right]$ $\left[\because a^{3}+b^{3}=(a+b)\left(a^{2}+b^{2}-a b\right)\right]$ $=(3 y+5 z)\left[9 y^{2}+25 z^{2}-15 y z\right]$ (ii) $64 m^{3}-343 n^{3}$ $=(4 m)^{3}-(7 n)^{3}$ $=(4 m-7 n)\left[(4 m)^{2}+(7 n)^{2}+(4 m)(7 n)\right]$ $\left[\because a^{3}-b^{3}=(a-b)\left(a^{2}+b^{2}+a b\right)\right]$ $=(4 m-7 n)\left[16 m^{2}+49 n^{2}+28 m n\right]$...
Read More →Verify: <br/><br/>(i) $x^{3}+y^{3}=(x+y)\left(x^{2}-x y+y^{2}\right)$<br/><br/>(ii) $x^{3}-y^{3}=(x-y)\left(x^{2}+x y+y^{2}\right)$
Solution: (i) It is known that, $(x+y)^{3}=x^{3}+y^{3}+3 x y(x+y)$ $x^{3}+y^{3}=(x+y)^{3}-3 x y(x+y)$ $=(x+y)\left[(x+y)^{2}-3 x y\right]$ $=(x+y)\left(x^{2}+y^{2}+2 x y-3 x y\right)$ $=(x+y)\left(x^{2}+y^{2}-x y\right)$ $=(x+y)\left(x^{2}-x y+y^{2}\right)$ (ii) It is known that, $(x-y)^{3}=x^{3}-y^{3}-3 x y(x-y)$ $x^{3}-y^{3}=(x-y)^{3}+3 x y(x-y)$ $=(x-y)\left[(x-y)^{2}+3 x y\right]$ $=(x-y)\left(x^{2}+y^{2}-2 x y+3 x y\right)$ $=(x-y)\left(x^{2}+y^{2}+x y\right)$ $=(x-y)\left(x^{2}+x y+y^{2}\r...
Read More →Evaluate the following using suitable identities: <br/><br/>(i) $(99)^{3}$ <br/><br/>(ii) $(102)^{3}$ <br/><br/>(iii) $(998)^{3}$
Solution: It is known that, $(a+b)^{3}=a^{3}+b^{3}+3 a b(a+b)$ and $(a-b)^{3}=a^{3}-b^{3}-3 a b(a-b)$ (i) $(99)^{3}=(100-1)^{3}$ $=(100)^{3}-(1)^{3}-3(100)(1)(100-1)$ $=1000000-1-300(99)$ $=1000000-1-29700$ $=970299$ (ii) $(102)^{3}=(100+2)^{3}$ $=(100)^{3}+(2)^{3}+3(100)(2)(100+2)$ $=1000000+8+600(102)$ $=1000000+8+61200$ $=1061208$ (iii) $(998)^{3}=(1000-2)^{3}$ $=(1000)^{3}-(2)^{3}-3(1000)(2)(1000-2)$ $=1000000000-8-6000(998)$ $=1000000000-8-5988000$ $=1000000000-5988008$ $=994011992$...
Read More →Write the following cubes in expanded form: <br/><br/>(i) $(2 x+1)^{3}$ <br/><br/>(ii) $(2 a-3 b)^{3}$<br/><br/>(iii) $\left[\frac{3}{2} x+1\right]^{3}$ <br/><br/>(iv) $\left[x-\frac{2}{3} y\right]^{3}$
Solution: It is known that, $(a+b)^{3}=a^{3}+b^{3}+3 a b(a+b)$ and $(a-b)^{3}=a^{3}-b^{3}-3 a b(a-b)$ (i) $(2 x+1)^{3}=(2 x)^{3}+(1)^{3}+3(2 x)(1)(2 x+1)$ $=8 x^{3}+1+6 x(2 x+1)$ $=8 x^{3}+1+12 x^{2}+6 x$ $=8 x^{3}+12 x^{2}+6 x+1$ (ii) $(2 a-3 b)^{3}=(2 a)^{3}-(3 b)^{3}-3(2 a)(3 b)(2 a-3 b)$ $=8 a^{3}-27 b^{3}-18 a b(2 a-3 b)$ $=8 a^{3}-27 b^{3}-36 a^{2} b+54 a b^{2}$ (iii) $\left[\frac{3}{2} x+1\right]^{3}=\left[\frac{3}{2} x\right]^{3}+(1)^{3}+3\left(\frac{3}{2} x\right)(1)\left(\frac{3}{2} x+...
Read More →Factorise: <br/> <br/>(i) $4 x^{2}+9 y^{2}+16 z^{2}+12 x y-24 y z-16 x z$<br/> <br/>(ii) $2 x^{2}+y^{2}+8 z^{2}-2 \sqrt{2} x y+4 \sqrt{2} y z-8 x z$
Solution: It is known that, $(x+y+z)^{2}=x^{2}+y^{2}+z^{2}+2 x y+2 y z+2 z x$ (i) $4 x^{2}+9 y^{2}+16 z^{2}+12 x y-24 y z-16 x z$ $=(2 x)^{2}+(3 y)^{2}+(-4 z)^{2}+2(2 x)(3 y)+2(3 y)(-4 z)+2(2 x)(-4 z)$ $=(2 x+3 y-4 z)^{2}$ $=(2 x+3 y-4 z)(2 x+3 y-4 z)$ (ii) $2 x^{2}+y^{2}+8 z^{2}-2 \sqrt{2} x y+4 \sqrt{2} y z-8 x z$ $=(-\sqrt{2} x)^{2}+(y)^{2}+(2 \sqrt{2} z)^{2}+2(-\sqrt{2} x)(y)+2(y)(2 \sqrt{2} z)+2(-\sqrt{2} x)(2 \sqrt{2} z)$ $=(-\sqrt{2} x+y+2 \sqrt{2} z)^{2}$ $=(-\sqrt{2} x+y+2 \sqrt{2} z)(-...
Read More →Factorise the following using appropriate identities: <br/><br/>(i) $9 x^{2}+6 x y+y^{2}$<br/><br/> (ii) $4 y^{2}-4 y+1$ <br/><br/>(iii) $x^{2}-\frac{y^{2}}{100}$
Solution: (i) $9 x^{2}+6 x y+y^{2}=(3 x)^{2}+2(3 x)(y)+(y)^{2}$ $=(3 x+y)(3 x+y)$ $\left[x^{2}+2 x y+y^{2}=(x+y)^{2}\right]$ (ii) $4 y^{2}-4 y+1=(2 y)^{2}-2(2 y)(1)+(1)^{2}$ $=(2 y-1)(2 y-1)$ $\left[x^{2}-2 x y+y^{2}=(x-y)^{2}\right]$ (iii) $x^{2}-\frac{y^{2}}{100}=x^{2}-\left(\frac{y}{10}\right)^{2}$ $=\left(x+\frac{y}{10}\right)\left(x-\frac{y}{10}\right)$ $\left[x^{2}-y^{2}=(x+y)(x-y)\right]$...
Read More →Evaluate the following products without multiplying directly: <br/> <br/> (i) $103 \times 107$ <br/> <br/>(ii) $95 \times 96$<br/> <br/> (iii) $104 \times 96$
Solution: (i) $103 \times 107=(100+3)(100+7)$ $=(100)^{2}+(3+7) 100+(3)(7)$ [By using the identity $(x+a)(x+b)=x^{2}+(a+b) x+a b$, where $x=100, a=3$, and $b=7]$ $=10000+1000+21$ $=11021$ (ii) $95 \times 96=(100-5)(100-4)$ $=(100)^{2}+(-5-4) 100+(-5)(-4)$ [By using the identity $(x+a)(x+b)=x^{2}+(a+b) x+a b$, where $x=100, a=-5$, and $b=-4]$ $=10000-900+20$ $=9120$ (iii) $104 \times 96=(100+4)(100-4)$ $=(100)^{2}-(4)^{2}\left[(x+y)(x-y)=x^{2}-y^{2}\right]$ $=10000-16$ $=9984$...
Read More →Factorise:<br/><br/>(i) $x^{3}-2 x^{2}-x+2$ <br/><br/>(ii) $x^{3}+3 x^{2}-9 x-5$<br/><br/>(iii) $x^{3}+13 x^{2}+32 x+20$<br/><br/> (iv) $2 y^{3}+y^{2}-2 y-1$
Solution: (i) Let $p(x)=x^{3}-2 x^{2}-x+2$ All the factors of 2 have to be considered. These are $\pm 1, \pm 2$ By trial method $p(-1)=(-1)^{3}-2(-1)^{2}-(-1)+2$ $=-1-2+1+2=0$ Therefore, $(x+1)$ is factor of polynomial $p(x)$. Let us find the quotient on dividing $x^{3}-2 x^{2}-x+2$ by $x+1$. By long division, It is known that, Dividend $=$ Divisor $\times$ Quotient $+$ Remainder $\therefore x^{3}-2 x^{2}-x+2=(x+1)\left(x^{2}-3 x+2\right)+0$ $=(x+1)\left[x^{2}-2 x-x+2\right]$ $=(x+1)[x(x-2)-1(x-...
Read More →Factorise: <br/> <br/> (i) $12 x^{2}-7 x+1$<br/> <br/> (ii) $2 x^{2}+7 x+3$<br/> <br/>(iii) $6 x^{2}+5 x-6$ <br/> <br/>(iv) $3 x^{2}-x-4$
Solution: (i) $12 x^{2}-7 x+1$ We can find two numbers such that $p q=12 \times 1=12$ and $p+q=-7$. They are $p=-4$ and $q=-3$. Here, $12 x^{2}-7 x+1=12 x^{2}-4 x-3 x+1$ $=4 x(3 x-1)-1(3 x-1)$ $=(3 x-1)(4 x-1)$ (ii) $2 x^{2}+7 x+3$ We can find two numbers such that $p q=2 \times 3=6$ and $p+q=7$. They are $p=6$ and $q=1$. Here, $2 x^{2}+7 x+3=2 x^{2}+6 x+x+3$ $=2 x(x+3)+1(x+3)$ $=(x+3)(2 x+1)$ (iii) $6 x^{2}+5 x-6$ We can find two numbers such that $p q=-36$ and $p+q=5$. They are $p=9$ and $q=-4...
Read More →Find the value of k, if x − 1 is a factor of p(x) in each of the following cases: <br/> <br/> (i) $p(x)=x^{2}+x+k$ <br/> <br/>(ii) $p(x)=2 x^{2}+k x+\sqrt{2}$ <br/> <br/>(iii) $p(x)=k x^{2}-\sqrt{2} x+1$<br/> <br/> (iv) $p(x)=k x^{2}-3 x+k$
Solution: If $x$ - 1 is a factor of polynomial $p(x)$, then $p(1)$ must be 0 . (i) $p(x)=x^{2}+x+k$ $p(1)=0$ $\Rightarrow(1)^{2}+1+k=0$ $\Rightarrow 2+k=0$ $\Rightarrow k=-2$ Therefore, the value of $k$ is $-2$. (ii) $p(x)=2 x^{2}+k x+\sqrt{2}$ $p(1)=0$ $\Rightarrow 2(1)^{2}+k(1)+\sqrt{2}=0$ $\Rightarrow 2+k+\sqrt{2}=0$ $\Rightarrow k=-2-\sqrt{2}=-(2+\sqrt{2})$ Therefore, the value of $k$ is $-(2+\sqrt{2})$. (iii) $p(x)=k x^{2}-\sqrt{2} x+1$ $p(1)=0$ $\Rightarrow k(1)^{2}-\sqrt{2}(1)+1=0$ $\Righ...
Read More →Determine which of the following polynomials has (x + 1) a factor: <br/> <br/>(i) $x^{3}+x^{2}+x+1$<br/> <br/> (ii) $x^{4}+x^{3}+x^{2}+x+1$<br/> <br/> (iii) $x^{4}+3 x^{3}+3 x^{2}+x+1$ <br/> <br/>(iv) $x^{3}-x^{2}-(2+\sqrt{2}) x+\sqrt{2}$
Solution: (i) If $(x+1)$ is a factor of $p(x)=x^{3}+x^{2}+x+1$, then $p(-1)$ must be zero, otherwise $(x+1)$ is not a factor of $p(x)$. $p(x)=x^{3}+x^{2}+x+1$ $p(-1)=(-1)^{3}+(-1)^{2}+(-1)+1$ $=-1+1-1+1=0$ Hence, $x+1$ is a factor of this polynomial. (ii) If $(x+1)$ is a factor of $p(x)=x^{4}+x^{3}+x^{2}+x+1$, then $p(-1)$ must be zero, otherwise $(x+1)$ is not a factor of $p(x)$. $p(x)=x^{4}+x^{3}+x^{2}+x+1$ $p(-1)=(-1)^{4}+(-1)^{3}+(-1)^{2}+(-1)+1$ $=1-1+1-1+1=1$ As $p(-1) \neq 0$ Therefore, $...
Read More →Check whether 7 + 3x is a factor of 3 x 3 + 7x.
Solution: Let us divide $\left(3 x^{3}+7 x\right)$ by $(7+3 x)$. If the remainder obtained is 0 , then $7+3 x$ will be a factor of $3 x^{3}+7 x$. By long division, As the remainder is not zero, therefore, $7+3 x$ is not a factor of $3 x^{3}+7 x$....
Read More →Find the remainder when $x^{3}-a x^{2}+6 x-a$ is divided by x − a.
Solution: By long division, Therefore, when $x^{3}-a x^{2}+6 x-a$ is divided by $x-a$, the remainder obtained is $5 a$....
Read More →Find the remainder when $x^{3}+3 x^{2}+3 x+1$ is divided by <br/> <br/>(i) $x+1$ <br/> <br/>(ii) $x-\frac{1}{2}$ <br/> <br/>(iii) $x$(iv) $x+\pi$ <br/> <br/>(v) $5+2 x$
Solution: (i) $x+1$ By long division, Therefore, the remainder is 0 . (ii) $x-\frac{1}{2}$ By long division, Therefore, the remainder is $\frac{27}{8}$. (iii) $X$ By long division, Therefore, the remainder is 1 . (iv) $x+\pi$ By long division, Therefore, the remainder is $-\pi^{3}+3 \pi^{2}-3 \pi+1$. (v) $5+2 x$ By long division, Therefore, the remainder is $-\frac{27}{8}$...
Read More →Find p(0), p(1) and p(2) for each of the following polynomials: <br/> <br/> (i) $p(y)=y^{2}-y+1$<br/> <br/> (ii) $p(t)=2+t+2 t^{2}-t^{3}$<br/> <br/>(iii) $p(x)=x^{3}$ <br/> <br/>(iv) $p(x)=(x-1)(x+1)$
Solution: (i) $p(y)=y^{2}-y+1$ $p(0)=(0)^{2}-(0)+1=1$ $p(1)=(1)^{2}-(1)+1=1$ $p(2)=(2)^{2}-(2)+1=3$ (ii) $p(t)=2+t+2 t^{2}-t^{3}$ $p(0)=2+0+2(0)^{2}-(0)^{3}=2$ $p(1)=2+(1)+2(1)^{2}-(1)^{3}$ $=2+1+2-1=4$ $p(2)=2+2+2(2)^{2}-(2)^{3}$ $=2+2+8-8=4$ (iii) $p(x)=x^{3}$ $p(0)=(0)^{3}=0$ $p(1)=(1)^{3}=1$ $p(2)=(2)^{3}=8$ (iv) $p(x)=(x-1)(x+1)$ $p(0)=(0-1)(0+1)=(-1)(1)=-1$ $p(1)=(1-1)(1+1)=0(2)=0$ $p(2)=(2-1)(2+1)=1(3)=3$...
Read More →Find the value of the polynomial $5 x-4 x^{2}+3$ at <br/> <br/>(i) $x=0$ <br/> <br/>(ii) $x=-1$ <br/> <br/>(iii) $x=2$
Solution: (i) $p(x)=5 x-4 x^{2}+3$ $p(0)=5(0)-4(0)^{2}+3$ $=3$ (ii) $p(x)=5 x-4 x^{2}+3$ $p(-1)=5(-1)-4(-1)^{2}+3$ $=-5-4(1)+3=-6$ (iii) $p(x)=5 x-4 x^{2}+3$ $p(2)=5(2)-4(2)^{2}+3$ $=10-16+3=-3$...
Read More →Classify the following as linear, quadratic and cubic polynomial: <br/> <br/> (i) $x^{2}+x$ <br/> <br/>(ii) $x-x^{3}$<br/> <br/>(iii) $y+y^{2}+4$<br/> <br/>(iv) $1+x$ <br/> <br/>(v) $3 t$<br/> <br/>(vi) $r^{2}$<br/> <br/>(vii) $7 x^{3}$
Solution: Linear polynomial, quadratic polynomial, and cubic polynomial has its degrees as 1,2 , and 3 respectively. (i) $x^{2}+x$ is a quadratic polynomial as its degree is $2 .$ (ii) $x-x^{3}$ is a cubic polynomial as its degree is 3 . (iii) $y+y^{2}+4$ is a quadratic polynomial as its degree is 2 . (iv) $1+x$ is a linear polynomial as its degree is 1 . (v) $3 t$ is a linear polynomial as its degree is 1 . (vi) $r^{2}$ is a quadratic polynomial as its degree is 2 . (vii) $7 x^{3}$ is a cubic p...
Read More →Write the degree of each of the following polynomials: <br/> <br/> (i) $5 x^{3}+4 x^{2}+7 x$ <br/> <br/>(ii) $4-y^{2}$ <br/> <br/>(iii) $5 t-\sqrt{7}$ <br/> <br/>(iv) 3
Solution: Degree of a polynomial is the highest power of the variable in the polynomial. (i) $5 x^{3}+4 x^{2}+7 x$ This is a polynomial in variable $x$ and the highest power of variable $x$ is 3 . Therefore, the degree of this polynomial is $3 .$ (ii) $4-y^{2}$ This is a polynomial in variable $y$ and the highest power of variable $y$ is 2 . Therefore, the degree of this polynomial is 2 . (iii) $5 t-\sqrt{7}$ This is a polynomial in variable $t$ and the highest power of variable $t$ is 1 . There...
Read More →Give one example each of a binomial of degree 35, and of a monomial of degree 100.
Solution: Degree of a polynomial is the highest power of the variable in the polynomial. Binomial has two terms in it. Therefore, binomial of degree 35 can be written as $x^{35}+x^{34}$. Monomial has only one term in it. Therefore, monomial of degree 100 can be written as $x^{100}$....
Read More →Write the coefficients of $x^{2}$ in each of the following: <br/> <br/> (i) $2+x^{2}+x$ <br/> <br/>(ii) $2-x^{2}+x^{3}$ <br/> <br/>(iii) $\frac{\pi}{2} x^{2}+x$ <br/> <br/>(iv) $\sqrt{2} x-1$
Solution: (i) $2+x^{2}+x$ In the above expression, the coefficient of $x^{2}$ is 1 . (ii) $2-x^{2}+x^{3}$ In the above expression, the coefficient of $x^{2}$ is $-1$. (iii) $\frac{\pi}{2} x^{2}+x$ In the above expression, the coefficient of $x^{2}$ is $\frac{\pi}{2}$. (iv) $\sqrt{2} x-1$, or $0 x^{2}+\sqrt{2} x-1$ In the above expression, the coefficient of $x^{2}$ is $0 .$...
Read More →Simplify: <br/><br/>(i)$2^{\frac{2}{3}} \cdot 2^{\frac{1}{5}}$ <br/><br/>(ii)$\left(\frac{1}{3^{3}}\right)^{7}$ <br/><br/> (iii) $\frac{11^{\frac{1}{2}}}{11^{\frac{1}{4}}}$ <br/><br/> (iv) $7^{\frac{1}{2}} \cdot 8^{\frac{1}{2}}$
Solution: (i) $2^{\frac{2}{3}} \cdot 2^{\frac{1}{5}}=2^{\frac{2}{3}+\frac{1}{5}}$ $\left[a^{m} \cdot a^{n}=a^{m+n}\right]$ $=2^{\frac{10+3}{15}}=2^{\frac{13}{15}}$ (ii)$\left(\frac{1}{3^{3}}\right)^{7}=\frac{1}{3^{3 \times 7}}$ $\left[\left(a^{m}\right)^{n}=a^{m m}\right]$ $=\frac{1}{3^{21}}$ $=3^{-21}$ $\left[\frac{1}{a^{m}}=a^{-m}\right]$ (iii) $\frac{11^{\frac{1}{2}}}{11^{\frac{1}{4}}}=11^{\frac{1}{2}-\frac{1}{4}}$ $\left[\frac{a^{m}}{a^{n}}=a^{m-n}\right]$ $=11^{\frac{2-1}{4}}=11^{\frac{1}{4...
Read More →Find: <br/> <br/> (i)$9^{\frac{3}{2}}$<br/> <br/> (ii)$32^{\frac{2}{5}}$ <br/> <br/>(iii)$16^{\frac{3}{4}}$<br/> <br/> (iv)$125^{\frac{-1}{3}}$
Solution: (i) $9^{\frac{3}{2}}=\left(3^{2}\right)^{\frac{3}{2}}$ $=3^{2 \times \frac{3}{2}}$ $\left[\left(a^{m}\right)^{n}=a^{m m}\right]$ $=3^{3}=27$ (ii)$(32)^{\frac{2}{5}}=\left(2^{5}\right)^{\frac{2}{5}}$ $=2^{5 \times \frac{2}{5}}$ $\left[\left(a^{m}\right)^{n}=a^{m n}\right]$ $=2^{2}=4$ (iii) $(16)^{\frac{3}{4}}=\left(2^{4}\right)^{\frac{3}{4}}$ $=2^{4 \times \frac{3}{4}}$ $\left[\left(a^{m}\right)^{n}=a^{m n}\right]$ $=2^{3}=8$ (iv)$(125)^{\frac{-1}{3}}=\frac{1}{(125)^{\frac{1}{3}}}$ $\le...
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