What must be subtracted from the polynomial f(x) = x4 + 2x3 − 13x2 − 12x + 21
Question: What must be subtracted from the polynomial $f(x)=x^{4}+2 x^{3}-13 x^{2}-12 x+21$ so that the resulting polynomial is exactly divisible by $x^{2}-4 x+3 ?$ Solution: We know that Dividend $=$ Quotient $\times$ Divisor $+$ Remainder . Dividend $-$ Remainder $=$ Quotient $\times$ Divisor . Clearly, Right hand side of the above result is divisible by the divisor. Therefore, left hand side is also divisible by the divisor. Thus, if we subtract remainder from the dividend, then it will be ex...
Read More →What must be added to the polynomial f(x) = x4 + 2x3 − 2x2 + x − 1 so
Question: What must be added to the polynomial $f(x)=x^{4}+2 x^{3}-2 x^{2}+x-1$ so that the resulting polynomial is exactly divisible by $x^{2}+2 x-3 ?$ Solution: We know that, $f(x)=g(x) \times q(x)+r(x)$ $f(x)-r(x)=g(x) \times q(x)$ $f(x)+\{-r(x)\}=g(x) \times q(x)$ Clearly, Right hand side is divisible by $g(x)$. Therefore, Left hand side is also divisible by $g(x)$. Thus, if we add $-r(x)$ to $f(x)$, then the resulting polynomial is divisible by $g(x)$. Let us now find the remainder when $f(...
Read More →Find all zeros of the polynomial f(x) = 2x4 − 2x3 − 7x2 + 3x + 6,
Question: Find all zeros of the polynomial $f(x)=2 x^{4}-2 x^{3}-7 x^{2}+3 x+6$, if its two zeros are $-\sqrt{\frac{3}{2}}$ and $\sqrt{\frac{3}{2}}$. Solution: Since $-\sqrt{\frac{3}{2}}$ and $\sqrt{\frac{3}{2}}$ are two zeros of $f(x)$. Therefore, $=\left(x-\sqrt{\frac{3}{2}}\right)\left(x+\sqrt{\frac{3}{2}}\right)$ $=\left(x^{2}-\frac{3}{2}\right)$ $=\frac{1}{2}\left(2 x^{2}-3\right)$ is a factor of $f(x)$. Also $2 x^{2}-3$ is a factor of $f(x)$. Let us now divide $f(x)$ by $2 x^{2}-3$. We hav...
Read More →Rationalize the denominator and simplify:
Question: Rationalize the denominator and simplify: (i) $\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}$ (ii) $\frac{5+2 \sqrt{3}}{7+4 \sqrt{3}}$ (iii) $\frac{1+\sqrt{2}}{3-2 \sqrt{2}}$ (iv) $\frac{2 \sqrt{6}-\sqrt{5}}{3 \sqrt{5}-2 \sqrt{6}}$ (v) $\frac{4 \sqrt{3}+5 \sqrt{2}}{\sqrt{48}+\sqrt{18}}$ (vi) $\frac{2 \sqrt{3}-\sqrt{5}}{2 \sqrt{2}+3 \sqrt{3}}$ Solution: (i) $\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}$ Rationalizing the denominator by multiplying both numerator and denominator with the ra...
Read More →Obtain all zeros of the polynomial f(x) = x4 − 3x3 − x2 + 9x − 6,
Question: Obtain all zeros of the polynomial $1(x)=x^{4}-3 x^{3}-x^{2}+9 x-6$, if two of its zeros are $-\sqrt{3}$ and $\sqrt{3}$ Solution: We know that if $x=\alpha$ is a zero of a polynomial, then $(x-\alpha)$ is a factor of $f(x)$. Since $-\sqrt{3}$ and $\sqrt{3}$ are zeros of $f(x)$. Therefore $(x+\sqrt{3})(x-\sqrt{3})=x^{2}+\sqrt{3} x-\sqrt{3} x-3$ $=x^{2}-3$ $x^{2}-3$ is a factor of $f(x)$. Now, we divide $f(x)=x^{4}-3 x^{3}-x^{2}+9 x-6$ by $g(x)=x^{2}-3$ to find the other zeros of $f(x)$....
Read More →Obtain all zeros of f(x) = x3 + 13x2 + 32x + 20,
Question: Obtain all zeros of $1(x)=x^{3}+13 x^{2}+32 x+20$, if one of its zeros is $-2$. Solution: Since $-2$ is one zero of $f(x)$. Therefore, we know that, if $x=\alpha$ is a zero of a polynomial, then $(x-\alpha)$ is a factor of $f(x)=x+2$ is a factor of $f(x)$. Now, we divide $f(x)=x^{3}+13 x^{2}+32 x+20$ by $g(x)=(x+2)$ to find the others zeros of $f(x)$. By using that division algorithm we have, $f(x)=g(x) \times q(x)+r(x)$ $x^{3}+13 x^{2}+32 x+20=(x+2)\left(x^{2}+11 x+10\right)+0$ $x^{3}...
Read More →Express each one of the following with rational denominator:
Question: Express each one of the following with rational denominator: (i) $\frac{1}{3+\sqrt{2}}$ (ii) $\frac{1}{\sqrt{6}-\sqrt{5}}$ (iii) $\frac{16}{\sqrt{41}-5}$ (iv) $\frac{30}{5 \sqrt{3}-3 \sqrt{5}}$ (v) $\frac{1}{2 \sqrt{5}-\sqrt{3}}$ (vi) $\frac{\sqrt{3}+1}{2 \sqrt{2}-\sqrt{3}}$ (vii) $\frac{6-4 \sqrt{2}}{6+4 \sqrt{2}}$ (viii) $\frac{3 \sqrt{2}+1}{2 \sqrt{5}-3}$ (ix) $\frac{b^{2}}{\sqrt{\left(a^{2}+b^{2}\right)}+a}$ Solution: (i) $\frac{1}{3+\sqrt{2}}$ Rationalizing the denominator by mult...
Read More →Obtain all zeros of the polynomial f(x) = 2x4 + x3 − 14x2 − 19x − 6, if two of its zeros are −2 and −1.
Question: Obtain all zeros of the polynomial $1(x)=2 x^{4}+x^{3}-14 x^{2}-19 x-6$, if two of its zeros are $-2$ and $-1$. Solution: We know that, if $x=\alpha$ is a zero of a polynomial, and then $x-\alpha$ is a factor of $f(x)$ Since $_{-2}$ and $_{-1}$ are zeros of $f(x)$. Therefore $(x+2)(x+1)=x^{2}+2 x+x+2$ $=x^{2}+3 x+2$ $x^{2}+3 x+2$ is a factor of $f(x)$. Now, We divide $2 x^{4}+x^{3}-14 x^{2}-19 x-6$ by $g(x)=x^{2}+3 x+2$ to find the other zeros of $f(x)$. By using division algorithm we ...
Read More →Check whether the first polynomial is a factor of the second polynomial by applying the division algorithm :
Question: Check whether the first polynomial is a factor of the second polynomial by applying the division algorithm : (i) $g(t)=t^{2}-3, f(t)=2 t^{4}+3 t^{3}-2 t^{2}-9 t-12$ (ii) $g(x)=x^{3}-3 x+1, f(x)=x^{5}-4 x^{3}+x^{2}+3 x+1$ (iii) $g(x)=2 x^{2}-x+3, f(x)=6 x^{5}-x^{4}+4 x^{3}-5 x^{2}-x-15$ Solution: $(i)$. Given $g(t)=t^{2}-3$ $f(t)=2 t^{4}+3 t^{3}-2 t^{2}-9 t-12$ Here, degree $(f(t))=4$ and Degree $(g(t))=2$ Therefore, quotient $q(t)$ is of degree $4-2=2$ Remainder $r(t)$ is of degree 1 o...
Read More →Find the value to three places of decimals of each of the following. It is given that
Question: Find the value to three places of decimals of each of the following. It is given that $\sqrt{2}=1.414, \sqrt{3}=1.732, \sqrt{5}=2.236, \sqrt{10}=3.162 .$ (i) $\frac{2}{\sqrt{3}}$ (ii) $\frac{3}{\sqrt{10}}$ (iii) $\frac{\sqrt{5}+1}{\sqrt{2}}$ (iv) $\frac{\sqrt{10}+\sqrt{15}}{\sqrt{2}}$ (v) $\frac{2+\sqrt{3}}{3}$ (vi) $\frac{\sqrt{2}-1}{\sqrt{5}}$ Solution: Given, $\sqrt{2}=1.414, \sqrt{3}=1.732, \sqrt{5}=2.236, \sqrt{10}=3.162$ (i) $\frac{2}{\sqrt{3}}$ Rationalizing the denominator by m...
Read More →Rationalize the denominator of each of the following:
Question: Rationalize the denominator of each of the following: (i) $\frac{3}{\sqrt{5}}$ (ii) $\frac{3}{2 \sqrt{5}}$ (iii) $\frac{1}{\sqrt{12}}$ (iv) $\frac{\sqrt{2}}{\sqrt{3}}$ (v) $\frac{\sqrt{2}+\sqrt{5}}{\sqrt{3}}$ (vi) $\frac{\sqrt{2}+\sqrt{5}}{\sqrt{3}}$ (vii) $\frac{3 \sqrt{2}}{5}$ Solution: (i) $\frac{3}{\sqrt{5}}$ For rationalizing the denominator, multiply both numerator and denominator with $\sqrt{5}$ $=\frac{3 \times \sqrt{5}}{\sqrt{5} \times \sqrt{5}}=\frac{3 \times \sqrt{5}}{\sqrt{...
Read More →Apply division algorithm to find the quotient
Question: Apply division algorithm to find the quotientq(x) and remainderr(x) in dividingf(x) byg(x) in each of the following : (i) $f(x)=x^{3}-6 x^{2}+11 x-6, g(x)=x^{2}+x+1$ (ii) $f(x)=10 x^{4}+17 x^{3}-62 x^{2}+30 x-3, g(x)=2 x^{2}+7 x+1$ (iii) $f(x)=4 x^{3}+8 x+8 x^{2}+7, g(x)=2 x^{2}-x+1$ (iv) $f(x)=15 x^{3}-20 x^{2}+13 x-12, g(x)=2-2 x+x^{2}$ Solution: (i) We have $f(x)=x^{3}-6 x^{2}+11 x-6$ $g(x)=x^{2}+x+1$ Here, degree $[f(x)]=3$ and Degree $(g(x))=2$ Therefore, quotient $q(x)$ is of deg...
Read More →Simplify the following expressions:
Question: Simplify the following expressions: (i) $(\sqrt{3}+\sqrt{7})^{2}$ (ii) $(\sqrt{5}-\sqrt{3})^{2}$ (iii) $(2 \sqrt{5}+3 \sqrt{2})^{2}$ Solution: (i) $(\sqrt{3}+\sqrt{7})^{2}$ As we know, $(a+b)^{2}=\left(a^{2}+2 \times a \times b+b^{2}\right)$ $=\sqrt{3^{2}}+2 \times \sqrt{3} \times \sqrt{7}+\sqrt{7^{2}}$ $=3+2 \times \sqrt{3 \times 7}+7$ $=10+2 \times \sqrt{21}$ (ii) $(\sqrt{5}-\sqrt{3})^{2}$ As we know, $(a-b)^{2}=\left(a^{2}-2 \times a \times b+b^{2}\right)$ $=(\sqrt{5})^{2}-2 \times ...
Read More →Simplify the following expressions:
Question: Simplify the following expressions: (i) $(11+\sqrt{11})(11-\sqrt{11})$ (ii) $(5+\sqrt{7})(5-\sqrt{7})$ (iii) $(\sqrt{8}-\sqrt{2})(\sqrt{8}+\sqrt{2})$ (iv) $(3+\sqrt{3})(3-\sqrt{3})$ (v) $(\sqrt{5}-\sqrt{2})(\sqrt{5}+\sqrt{2})$ Solution: (i) $(11+\sqrt{11})(11-\sqrt{11})$ As we know, $(a+b)(a-b)=\left(a^{2}-b^{2}\right)$ So, $11^{2}-11$ $121-11=110$ (ii) $(5+\sqrt{7})(5-\sqrt{7})$ As we know, $(a+b)(a-b)=\left(a^{2}-b^{2}\right)$ So, $5^{2}-7$ $25-7=18$ (iii) $(\sqrt{8}-\sqrt{2})(\sqrt{...
Read More →Simplify the following expressions:
Question: Simplify the following expressions: (i) $(4+\sqrt{7})(3+\sqrt{2})$ (ii) $(3+\sqrt{3})(5-\sqrt{2})$ (iii) $(\sqrt{5}-2)(\sqrt{3}-\sqrt{5})$ Solution: (i) $(4+\sqrt{7})(3+\sqrt{2})$ $=12+4 \sqrt{2}+3 \sqrt{7}+\sqrt{7 \times 2}$ $=12+4 \sqrt{2}+3 \sqrt{7}+\sqrt{14}$ (ii) $(3+\sqrt{3})(5-\sqrt{2})$ $=\sqrt{15}-3 \sqrt{2}+5 \sqrt{3}-\sqrt{3 \times 2}$ $=\sqrt{15}-3 \sqrt{2}+5 \sqrt{3}-\sqrt{6}$ (iii) $(\sqrt{5}-2)(\sqrt{3}-\sqrt{5})$ $=\sqrt{15}-\sqrt{25}-2 \sqrt{3}+2 \sqrt{5}$ $=\sqrt{15}-...
Read More →Simplify each of the following:
Question: Simplify each of the following: (i) $\sqrt[3]{4} \times \sqrt[3]{16}$ (ii) $\frac{\sqrt[4]{1250}}{\sqrt[4]{2}}$ Solution: (i) $\frac{\sqrt[4]{1250}}{\sqrt[4]{2}}$ (Note: $\sqrt[n]{a} \times \sqrt[n]{b}=\sqrt[n]{a \times b}$ ) $=\sqrt[3]{4 \times 16}$ $=\sqrt[3]{64}$ $=\sqrt[3]{4^{3}}$ $=\left(4^{3}\right)^{1 / 3}$ $=4^{(3 \times 1 / 3)}$ $=4^{1}$ $=4$ (ii) $\frac{\sqrt[4]{1250}}{\sqrt[4]{2}}$ $\left(\right.$ Note: $\left.\frac{\sqrt[n]{a}}{\sqrt[n]{b}}=\sqrt[n]{\frac{a}{b}}\right)$ $=\...
Read More →Show that
Question: Show that $\frac{\left(a+\frac{1}{b}\right)^{m} \times\left(a-\frac{1}{b}\right)^{n}}{\left(b+\frac{1}{a}\right)^{m} \times\left(b-\frac{1}{a}\right)^{n}}=\left(\frac{a}{b}\right)^{m+n}$ Solution: $=\frac{\left(\frac{a b+1}{b}\right)^{m} \times\left(\frac{a b-1}{b}\right)^{n}}{\left(\frac{a b+1}{a}\right)^{m} \times\left(\frac{a b+1}{a}\right)^{n}}$ $=\left(\frac{a}{b}\right)^{m} \times\left(\frac{a}{b}\right)^{n}$ $=\left(\frac{a}{b}\right)^{m+n}$ Hence, LHS = RHS...
Read More →Simplify
Question: Simplify (i) $\left(\frac{x^{a+b}}{x^{c}}\right)^{a-b}\left(\frac{x^{b+c}}{x^{a}}\right)^{b-c}\left(\frac{x^{c+a}}{x^{b}}\right)^{c-a}$ (ii) $\sqrt[m]{\frac{x^{1}}{x^{m}}} \times \sqrt[m n]{\frac{x^{m}}{x^{n}}} \times \sqrt[n]{\frac{x^{n}}{x^{1}}}$ Solution: (i) $\left(\frac{x^{a+b}}{x^{c}}\right)^{a-b}\left(\frac{x^{b+c}}{x^{a}}\right)^{b-c}\left(\frac{x^{c+a}}{x^{b}}\right)^{c-a}$ $\left(x^{a+b-c}\right)^{a-b}\left(x^{b+c-a}\right)^{b-c}\left(x^{c+a-b}\right)^{c-a}$ $\left(x^{a^{2}-b...
Read More →If the zeros of the polynomial f(x) = x3 − 12x2 + 39x + k are in A.P., find the value of k.
Question: If the zeros of the polynomial $t(x)=x^{3}-12 x^{2}+39 x+k$ are in A.P., find the value of $k$. Solution: Let $a-d, a$ and $a+d$ be the zeros of the polynomial $f(x)$ Then, Sum of the zeros $=\frac{\text { Coefficient of } x^{2}}{\text { Coefficient of } x^{3}}$ $a-d+a+a+d=\frac{-(-12)}{1}$ $3 a=12$ $a=\frac{12}{3}$ $a=4$ Since $a$ is a zero of the polynomial $f(x)$ $f(x)=x-12 x+39 x+k$ $f(a)=0$ $f(a)=4^{3}-12 \times 4^{2}+39 \times 4+k$ $0=64-192+156+k$ $0=220-192+k$ $0=28+k$ $-28=k$ ...
Read More →If the zeros of the polynomial f(x) = x3 − 12x2 + 39x + k are in A.P., find the value of k.
Question: If the zeros of the polynomial $1(x)=x^{3}-12 x^{2}+39 x+k$ are in A.P., find the value of $k$. Solution: Let $a-d, a$ and $a+d$ be the zeros of the polynomial $f(x)$. Then, Sum of the zeros $=\frac{\text { Coefficient of } x^{2}}{\text { Coefficient of } x^{3}}$ $a-d+a+a+d=\frac{-(-12)}{1}$ $a-\boldsymbol{A}+a+a+\boldsymbol{A}=12$ $3 a=12$ $a=\frac{12}{3}$ $a=4$ Since $a$ is a zero of the polynomial $f(x)$ $f(x)=x-12 x+39 x+k$ $f(a)=0$ $f(a)=4^{3}-12 \times 4^{2}+39 \times 4+k$ $0=64-...
Read More →If 1176 = 2a × 3b × 7c,
Question: If $1176=2 \times 3 \times 7$, find the values of $a, b$ and $c$. Solution: Hence compute the value of $2^{a} \times 3^{b} \times 7^{-c}$ as a fraction $1176=2^{a} \times 3^{b} \times 7^{c}$ $2^{3} \times 3^{1} \times 7^{2}=2^{a} \times 3^{b} \times 7^{c}$ $a=3, b=1, c=2$ We have to find the value of $2^{a} \times 3^{b} \times 7^{-c}$ $2^{a} \times 3^{b} \times 7^{-c}=2^{3} \times 3^{1} \times 7^{-2}$ $=\frac{2 \times 2 \times 2 \times 3}{7 \times 7}$ $=24 / 49$...
Read More →If 2x × 3y × 5z = 2160,
Question: If $2^{x} \times 3^{y} \times 5^{y}=2160$, find $x, y$ and $z$. Hence compute the value of $3^{x} \times 2^{-y} \times 5^{-z}$ Solution: $2^{x} \times 3^{y} \times 5^{z}=2160$ $2^{x} \times 3^{y} \times 5^{z}=2^{4} \times 3^{3} \times 5^{1}$ $x=4, y=3, z=1$ $3^{x} \times 2^{-y} \times 5^{-z}=3^{4} \times 2^{-3} \times 5^{-1}$ $=\frac{3 \times 3 \times 3 \times 3}{2 \times 2 \times 2 \times 5}$ $=81 / 40$...
Read More →The point on the curve
Question: The point on the curve $x^{2}=2 y$ which is nearest to the point $(0,5)$ is (A) $(2 \sqrt{2}, 4)$ (B) $(2 \sqrt{2}, 0)$ (C) $(0,0)$ (D) $(2,2)$ Solution: The given curve is $x^{2}=2 y$. For each value of $x$, the position of the point will be $\left(x, \frac{x^{2}}{2}\right)$. Let $\mathrm{P}\left(x, \frac{x^{2}}{2}\right)$ and $\mathrm{A}(0,5)$ are the given points. Now distance between the points $\mathrm{P}$ and $\mathrm{A}$ is given by, $\begin{aligned} \mathrm{PA} =\sqrt{(x-0)^{2}...
Read More →If a and b are different positive primes such that
Question: If a and b are different positive primes such that (i) $\left(\frac{a^{-1} b^{2}}{a^{2} b^{-4}}\right)^{7} \div\left(\frac{a^{3} b^{-5}}{a^{-2} b^{3}}\right)=a^{x} b^{y}$ find $x$ and $y$ (ii) $(a+b)^{-1}\left(a^{-1}+b^{-1}\right)=a^{x} b^{y}$, find $x$ and $y$ Solution: (i) $\left(\frac{a^{-1} b^{2}}{a^{2} b^{-4}}\right)^{7} \div\left(\frac{a^{3} b^{-5}}{a^{-2} b^{3}}\right)=a^{x} b^{y}$ find $x$ and $y$ $\left(\frac{a^{-1} b^{2}}{a^{2} b^{-4}}\right)^{7} \div\left(\frac{a^{3} b^{-5}}...
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