SSC > Polynomials

Explanation

{tex} 4x^{2}- \left(x^{2}-3\right)^{2} = \left(2x\right)^{2}- \left(x^{2}-3\right)^{2} {/tex}

= {tex} \left(2x+x^{2}-3\right) \left(2x-x^{2}+3\right) {/tex}

= {tex} \left(x^{2}+2x-3\right) \left(2x-x^{2}+3\right) {/tex}

= {tex} \left(x+3\right) \left(x-1\right) \left(3-x\right) \left(x+1\right) {/tex}

Explanation

When {tex} f \left(x\right)=2x^{3}-3x^{2}+4x-1 {/tex} is divided by {tex} x-1 {/tex}

Remainder = {tex} f \left(1\right) {/tex}

= {tex} 2.1^{3}-3.1^{2}+4.1-1 {/tex}

= 2

Explanation

Given : {tex} x+\frac{1}{3}=\sqrt{3} {/tex}

Taking cube on both sides, we have

{tex} \left(x+\frac{1}{x}\right)^{3}= \left(\sqrt{3}\right)^{3} {/tex}

=> {tex} x^{3} + \frac{1}{x^{3}} + 3x^{2}\frac{1}{x} + 3x\frac{1}{x^{2}} = 3\sqrt{3} {/tex}

=> {tex} x^{3}+\frac{1}{x^{3}}+3 \left(x+\frac{1}{x}\right)=3\sqrt{3} {/tex}

=> {tex} x^{3}+\frac{1}{x^{3}}+3 \times \sqrt{3} = 3\sqrt{3} {/tex}

=> {tex} x^{3}+\frac{1}{x^{3}} = 0 {/tex}

Explanation

{tex} f \left(x\right)=x^{3}+6x^{2}+11x+6 {/tex}

But x = -1 makes {tex} f \left(x\right)=0 {/tex}, therefore {tex} x+1 {/tex} must be factor of {tex} f \left(x\right) {/tex}.

Explanation

Since{tex} f(x) = x^{3} + 3x^{2} + Kx + 42 {/tex} is divisible by

{tex} x+3=x- \left(-3\right) {/tex}.

{tex} \therefore f\left(-3\right)=0 {/tex}

=> {tex} \left(-3\right)^{3}-3 \left(-3\right)^{2}+k \left(-3\right)+42 = 0 {/tex}

=> {tex} -27-27-3k+42 = 0 {/tex}

=> 3k + 12 = 0

=> k = -4

Explanation

{tex} x^{2}-4x = \left(x+2\right) \left(x-2\right) {/tex}

{tex} \therefore L.C.M.\ of\ \left(x-2\right)\ and\ \left(x^{2}-4\right) {/tex}

= {tex} 1 \times \left(x-2\right) \left(x+2\right) = x^{2}-4 {/tex}

and H.F.C. of {tex} \left(x-2\right)\ and\ \left(x^{2}-4\right) = x-2 {/tex}

Explanation

Given: {tex} a+b+c=0 {/tex}

=> {tex} a+b = -c {/tex}

Taking cube on both sides, we have

{tex} \left(a+b\right)^{3} = \left(-c\right)^{3} {/tex}

=> {tex} a^{3}+b^{3}+3ab \left(a+b\right)=-c^{3} {/tex}

=> {tex} a^{3}+b^{3}+3ab \left(-c\right)=-c^{3} {/tex}

=> {tex} a^{3}+b^{3}+c^{3}=3abc {/tex}

Explanation

Given: {tex} a^{m^{n}} = \left(a^{m}\right)^{n} = a^{m \times n} {/tex}

{tex} \therefore m^{n} = mn {/tex}

{tex} m^{n-1} = n {/tex}

{tex} \therefore m = n^{\frac{1}{n-1}} {/tex}

Explanation

{tex} \left(X^{b-c}\right)^{b+c}. \left(X^{c-a}\right)^{c+a}. \left(X^{a-b}\right)^{a+b} {/tex}

= {tex} \left(X\right)^{ \left(b-c\right) \left(b+c\right)}. \left(X\right)^{ \left(c-a\right) \left(c+a\right) }. \left(X\right)^{ \left(a-b\right) \left(a+b\right) } {/tex}

{tex} X^{b^{2}-c^{2}}.X^{c^{2}-a^{2}}.X^{a^{2}-b^{2}} {/tex}

= {tex} X^{b^{2}-c^{2}+c^{2}-a^{2}+a^{2}-b^{2}} {/tex}

={tex} X^{0} = 1 {/tex}

Explanation

Given : {tex} 2^{x} = 4^{y} = 8^{z} {/tex}

=> {tex} 2^{x} = 2^{2y} = 2^{3z} {/tex}

=> {tex} x = 2y = 3z = k\ (say) {/tex}

From {tex} \frac{1}{2x}+\frac{1}{4y}+\frac{1}{6z}=\frac{24}{7} {/tex}

{tex} \frac{1}{2k}+\frac{1}{2k}+\frac{1}{2k}=\frac{24}{7} {/tex}

=> {tex} \frac{3}{2k}=\frac{24}{7} {/tex}

=> {tex} k = \frac{7}{16} {/tex}

{tex} \therefore z = \frac{k}{3} = \frac{7}{16 \times 3} = \frac{7}{48} {/tex}

Explanation

{tex} X^{y} = Y^{x} {/tex}

{tex} \left(X\right)^{\frac{y}{x}} = Y {/tex}

{tex} \left(\frac{Y}{X}\right)^{\frac{x}{y}} = \frac{ \left(X^{\frac{y}{x}}\right)^{\frac{x}{y}} }{ \left(X\right)^{\frac{x}{y}} } {/tex}

={tex} \frac{X}{X^{\frac{x}{y}}} {/tex}

= {tex} \left(X\right)^{1-\frac{x}{y}} {/tex}

Explanation

Given: {tex} x+\frac{1}{x}=3 {/tex}

Taking cube on both sides, we have

{tex} \left(x+\frac{1}{x}\right)^{3}= \left(3\right)^{3} {/tex}

=> {tex} x^{3}+\frac{1}{x^{3}}+3. \left(x+\frac{1}{x}\right)=27 {/tex}

=> {tex} x^{3}+\frac{1}{x^{3}}+3 \times 3 = 27 {/tex}

=> {tex} x^{3}+\frac{1}{x^{3}} = 18 {/tex}

Explanation

If {tex} f \left(x\right) {/tex} is divided by {tex} x-3 {/tex}, then remainder will be {tex} f \left(3\right) {/tex}

Therefore, Remainder = {tex} f \left(3\right) {/tex}

={tex} 3^{5}-9 \times 3^{2}+12 \times 3-14 {/tex}

= {tex} 243 - 81 + 36 - 14 = 184 {/tex}

Explanation

When {tex} f \left(x\right) = x^{2}+ax+b {/tex} is divided by {tex} x-1 {/tex}, then remainder will be 5.

or {tex} f \left(1\right)=5 {/tex}

Therefore, 1 + a + b = 5

=> a + b = 4 ...(i)

When {tex} f \left(x\right) = x^{2}+ax+b {/tex} is divided by {tex} x+1 {/tex}, then remainder will be 5.

or {tex} f \left(-1\right)=5 {/tex}

Therefore, 1 - a + b = 5

=> -a + b = 4 ...(ii)

Solving equations (i) and (ii), we get

a = 0, b = 4

Explanation

Given polynomial is

{tex} x^{3}-6x^{2}+11x-6 {/tex}

{tex} Product\ of\ the\ zeros\ =

- \frac{constant\ term}{coefficient\ of\ x^{3}} {/tex}

= {tex} - \frac{ \left(-6\right) }{1} = 6 {/tex}

Explanation

{tex} x^{5}+2x^{4}+x^{3} = x^{3} \left(x^{2}+2x+1\right) = x^{3} \left(x+1\right)^{2} {/tex}

and {tex} x^{7}-x^{5} = x^{5} \left(x^{2}-1\right) {/tex}

= {tex} x^{5} \left(x+1\right) \left(x-1\right) {/tex}

{tex} \therefore H.C.F.\ =\ x^{3} \left(x+1\right) {/tex}

Explanation

Product of polynomials = H.C.F. X L.C.M.

Therefore, Other polynomials

= {tex} \frac{ \left(a-b\right) \times \left(a^{2}-b^{2}\right) \left(a^{2}+ab+b^{2}\right) }{a^{3}-b^{3}} {/tex}

= {tex} \frac{ \left(a-b\right) \times \left(a^{2}-b^{2}\right) \left(a^{2}+ab+b^{2}\right) }{ \left(a-b\right) \left(a^{2}+ab+b^{2}\right)} {/tex}

= {tex} a^{2} - b^{2} {/tex}

Explanation

Given polynomial

= {tex} 12x^{2}+11x+2 {/tex}

= {tex} 12x^{2}+5x+3x+2 {/tex}

= {tex} 4x \left(3x+2\right)+1 \left(3x+2\right) {/tex}

= {tex} \left(4x+1\right) \left(3x+2\right) {/tex}

Explanation

{tex} 2x^{2}+x-1 = 2x^{2}+2x-x-1 {/tex}

= {tex} 2x \left(x+1\right)-1 \left(x+1\right) {/tex}

= {tex} \left(2x-1\right) \left(x+1\right) {/tex} ...(i)

and {tex} 2x^{3}-3x^{2}-11x+6 {/tex}

= {tex} \left(2x-1\right) \left(x^{2}-x-6\right) {/tex}

= {tex} \left(2x-1\right) \left(x^{2}-3x+2x-6\right) {/tex}

= {tex} \left(2x-1\right)\left[ x \left(x-3\right)+2 \left(x-3\right) \right] {/tex}

= {tex} \left(2x-1\right) \left(x+2\right) \left(x-3\right) {/tex} ...(ii)

From equations (i) and (ii), we get

H.C.F. = {tex} \left(2x - 1\right) {/tex}

Explanation

{tex} x^{4}+x^{2}+1 = x^{4}+2x^{2}+1-x^{2} {/tex}
= {tex} \left(x^{2}+1\right)^{2}-x^{2} {/tex}
= {tex} \left(x^{2}+1+x\right) \left(x^{2}+1-x\right) {/tex} ...(i)
{tex} x^{4}-x^{2}-2x-1 = x^{4}- \left(x^{2}+2x+1\right) {/tex}
= {tex} \left(x^{2}\right)^{2}- \left(x+1\right)^{2} {/tex}
= {tex} \left(x^{2}+x+1\right) \left(x^{2}-x-1\right) {/tex} ...(ii)
{tex} x^{6}-1 = \left(x^{3}-1\right) \left(x^{3}+1\right) {/tex}
= {tex} \left(x-1\right) \left(x^{2}+x+1\right) \left(x+1\right) \left(x^{2}-x+1\right) {/tex} ...(iii)
L.C.M. = {tex} \left(x^{3}-1\right) \left(x^{3}+1\right) \left(x^{2}-x-1\right) {/tex}
= {tex} \left(x^{6}-1\right) \left(x^{2}-x-1\right) {/tex}

Explanation

{tex} 1681 = 41^{2} {/tex}