Q: If one zero of a quadratic polynomial \$x^2+3x+k\$ is \$2\$. Find the value of \$k\$.

\$2\$ is a zero of \$x^2+3x+k\$.Thus, \$(2)^2+3(2)+k=0\$\$k+10=0\$\$k=-10\$

Q: Find the zeroes of the polynomial \$x^{2} - 3\$ and verify the relationship between the zeroes and the coefficients.

Given polynomial is \$x^2 - 3\$ Here, \$a = 1, b= 0\$ and \$c = -3\$\$x^2 - 3 = (x - \sqrt{3})(x - \sqrt{3})\$So, the value of \$x^2 - 3\$ is zero when \$x = \sqrt{3}\$ or \$x = -\sqrt{3}\$Threrfore , thr zeroes of \$x^2 - 3\$ are \$\sqrt{3}\$ and \$-\sqrt{3}\$.Now, sum of zeroes \$= \sqrt{3} - \sqrt{3} = 0 = \dfrac{0}{1} = \dfrac{-b}{a}\$product of zeroes \$= (\sqrt{3} )(-\sqrt{3} )= -3 = \dfrac{-3}{1} = \dfrac{c}{a}\$Hence verified.

Q: Find \$p(0), p(1)\$ and \$p(2)\$ for each of the following polynomials:(i) \$p(y)=y^2-y+1\$(ii) \$p(t)=2+t+2t^2-t^3\$(iii) \$p(x)=x^3\$(iv) \$p(x)=(x-1)(x+1)\$

(i) \$p(y)=y^{2}-y+1\$\$p(0)=(0)^{2}-(0)+1=0+0+1=1\$\$p(1)=(1)^{2}-(1)+1=1-1+1=1\$\$p(2)=(2)^{2}-(2)+1=4-2+1=3\$(ii) \$p(t)=2+t+2t^{2}-t^{3}\$\$p(0)=2+(0)+2(0)^{2}-t^{3}=2+0+0-0=2\$\$p(1)=2+(1)+2(2)^{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\$

Q: Find the values of a and b so that \$x^4+x^3+8x^2+ax+b\$ is divisible by \$x^2+1\$.

Let us first divide the given polynomial \$x^4+x^3+8x^2+ax+b\$ by \$(x^2+1)\$ as shown in the above image:From the division, we observe that the quotient is \$x^2+x+7\$ and the remainder is \$(a-1)x+(b-7)\$.Since it is given that \$x^4+x^3+8x^2+ax+b\$ is exactly divisible by \$x^2+1\$, therefore, the remainder must be equal to \$0\$ that is:\$(a-1)x+(b-7)=0\\ \Rightarrow (a-1)x+(b-7)=0\cdot x+0\\ \Rightarrow (a-1)=0,\quad (b-7)=0\quad \quad \quad \quad (By\quad comparing\quad coefficients)\\ \Rightarrow a=1,\quad b=7\$Hence, \$a=1\$ and \$b=7\$.

Q: If two zeroes of the polynomial \$x^4- 6x^3 - 26x^2 + 138x -35\$ are \$2\pm \sqrt{3}\$, find the other zeroes.

The two zeroes of the polynomial is \$2+ \sqrt{3}, 2-\sqrt{3}\$Therefore, \$(x-2+ \sqrt{3})(x-2- \sqrt{3}) = x^2+4-4x-3\$= \$x^2-4x+1\$ is a factor of the given polynomial.Using division algorithm, we get\$x^4 -6x^3 -26x^2 + 138x -35 = (x^2-4x+1)(x^2-2x-35)\$So, \$(x^2-2x-35)\$ is also a factor of the given polynomial.\$x^2-2x-35 = x^2-7x+5x-35\$\$=x(x-7)+5(x-7)\$\$ = (x - 7)(x + 5)\$Hence, \$7\$ and \$-5 \$are the other zeros of this polynomial.

Q: Obtain all other zeroes of \$3x^4 +6x^3-2x^2-10x-5\$, if two of its zeroes are \$\sqrt{\dfrac{5}{3}}\$ and \$-\sqrt{\dfrac{5}{3}}\$.

Two zeroes are \$\sqrt{\dfrac{5}{3}}\$ and \$-\sqrt{\dfrac{5}{3}}\$So we can write it as, x = \$\sqrt{\dfrac{5}{3}}\$ and x = \$-\sqrt{\dfrac{5}{3}}\$we get \$x-\sqrt{\dfrac{5}{3}}=0\$ and \$x+\sqrt{\dfrac{5}{3}}=0\$Multiply both the factors we get,\$x^2-\dfrac{5}{3}=0\$Multiply by \$3\$ we get\$3x^2-5 = 0\$ is the factor of \$3x^4 +6x^3-2x^2-10x-5\$Now divide, \$3x^4 +6x^3-2x^2-10x-5\$ by \$3x^2-5 = 0\$ we get,Quotient is \$x^2+2x+1=0\$Compare the equation with quadratic formula,\$x^2 - (Sum \space\ of \space\ root)x + (Product \space\ of \space\ root) = 0\$\$\Rightarrow\$Sum of root \$= 2\$\$\Rightarrow\$Product of the root \$= 1\$So, we get\$\Rightarrow\$\$x^2+ x+x+1 = 0\$\$\Rightarrow\$\$x(x + 1) + 1(x + 1) = 0\$\$\Rightarrow\$\$x + 1 = 0, x + 1= 0\$\$\Rightarrow\$\$x = -1, x = -1\$So, our zeroes are \$-1, -1,\$ \$\sqrt{\dfrac{5}{3}}\$ and \$-\sqrt{\dfrac{5}{3}}\$

Q: Expand the following:\$(4a - b + 2c)^2\$

We have,\$(4a - b + 2c)^2 \\= (4a)^2 + (-b)^2 + (2c)^2 + 2(4a)(-b) + 2(-b)(2c) + 2(2c)(4a)\$\$\qquad[\because a^2 + b^2 + c^2 + 2ab + 2bc + 2ca = (a + b + c)^2 ]\$\$= 16a^2 + b^2 + 4c^2 - 8ab - 4ac + 16ca\$

Q: Simplify:\$(7m -8n)^2 + (7m + 8n)^2\$.

Given, \$(7m - 8n)^2 + (7m + 8n)^2\$.We know, \$(a-b)^2=a^2-2ab+b^2\$ and \$(a+b)^2=a^2+2ab+b^2\$.Then,\$(7m - 8n)^2 + (7m + 8n)^2\$\$=[(7m)^2-2(7m)(8n)+(8n)^2]+[(7m)^2+2(7m)(8n)+(8n)^2]\$\$=(49m^2-112mn+64n^2)+(49m^2+112mn+64n^2)\$\$=49m^2-112mn+64n^2+49m^2+112mn+64n^2\$\$=98m^2+128n^2\$.Therefore, option \$B\$ is correct.

Q: Show that: \$(4pq + 3q)^2 -(4pq -3q)^2 = 48pq^2\$

\$(4pq + 3q)^2 -(4pq -3q)^2 = 48pq^2\$LHS\$=(4pq)^2+2(4pq)(3q)+(3q)^2 - [(4pq)^2-2(4pq)(3q)+(3q)^2]\$ \$=16p^{ 2 }q^{ 2 }+24pq^{ 2 }+9q^2-16p^2q^2+24pq^2-9q^2\$ \$=48pq^2\$ =RHSLHS=RHS

Q: Show that: \$(9p - 5q)^2 + 180pq = (9p + 5q)^2\$

\$(9p - 5q)^2 + 180pq = (9p + 5q)^2\$LHS\$=(9p)^2-2(9p)(5q)+(5q)^2+180pq\$ \$= 81p^2 - 90pq + 25q^2 + 180pq\$ \$= 81p^2 + 90pq + 25q^2\$RHS\$= (9p + 5q)^2\$ \$=(9p)^2+2(9p)(5q)+(5q)^2\$ \$= 81p^2 + 90pq + 25q^2\$LHS=RHS

Question 1

If one zero of a quadratic polynomial $x^2+3x+k$ is $2$. Find the value of $k$.

Accepted Answer

$2$ is a zero of $x^2+3x+k$.Thus, $(2)^2+3(2)+k=0$$k+10=0$$k=-10$

Question 2

Find the zeroes of the polynomial $x^{2} - 3$ and verify the relationship between the zeroes and the coefficients.

Accepted Answer

Given polynomial is $x^2 - 3$ Here, $a = 1, b= 0$  and $c = -3$$x^2 - 3 = (x - \sqrt{3})(x - \sqrt{3})$So, the value of $x^2 - 3$ is zero when $x = \sqrt{3}$ or $x = -\sqrt{3}$Threrfore , thr zeroes of $x^2 - 3$  are $\sqrt{3}$  and $-\sqrt{3}$.Now, sum of zeroes $= \sqrt{3} - \sqrt{3} = 0 = \dfrac{0}{1} =  \dfrac{-b}{a}$product  of zeroes $= (\sqrt{3} )(-\sqrt{3} )= -3 = \dfrac{-3}{1} =  \dfrac{c}{a}$Hence verified.

Question 3

Find $p(0), p(1)$ and $p(2)$ for each of the following polynomials:(i) $p(y)=y^2-y+1$(ii) $p(t)=2+t+2t^2-t^3$(iii) $p(x)=x^3$(iv) $p(x)=(x-1)(x+1)$

Accepted Answer

(i) $p(y)=y^{2}-y+1$$p(0)=(0)^{2}-(0)+1=0+0+1=1$$p(1)=(1)^{2}-(1)+1=1-1+1=1$$p(2)=(2)^{2}-(2)+1=4-2+1=3$(ii) $p(t)=2+t+2t^{2}-t^{3}$$p(0)=2+(0)+2(0)^{2}-t^{3}=2+0+0-0=2$$p(1)=2+(1)+2(2)^{2}-(1)^{3}=2+1+2-1=4$$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$

Question 4

Find the values of a and b so that $x^4+x^3+8x^2+ax+b$ is divisible by $x^2+1$.

Accepted Answer

Let us first divide the given polynomial $x^4+x^3+8x^2+ax+b$ by $(x^2+1)$ as shown in the above image:From the division, we observe that the quotient is $x^2+x+7$ and the remainder is $(a-1)x+(b-7)$.Since it is given that $x^4+x^3+8x^2+ax+b$ is exactly divisible by $x^2+1$, therefore, the remainder must be equal to $0$ that is:$(a-1)x+(b-7)=0\ \Rightarrow (a-1)x+(b-7)=0\cdot x+0\ \Rightarrow (a-1)=0,\quad (b-7)=0\quad \quad \quad \quad (By\quad comparing\quad coefficients)\ \Rightarrow a=1,\quad b=7$Hence, $a=1$ and $b=7$.

Question 5

If two zeroes of the polynomial $x^4-  6x^3 - 26x^2 + 138x -35$ are $2\pm \sqrt{3}$, find the other zeroes.

Accepted Answer

The two zeroes of the polynomial is $2+ \sqrt{3}, 2-\sqrt{3}$Therefore, $(x-2+ \sqrt{3})(x-2- \sqrt{3}) = x^2+4-4x-3$= $x^2-4x+1$ is a factor of the given polynomial.Using division algorithm, we get$x^4 -6x^3 -26x^2 + 138x -35 = (x^2-4x+1)(x^2-2x-35)$So, $(x^2-2x-35)$ is also a factor of the given polynomial.$x^2-2x-35 = x^2-7x+5x-35$$=x(x-7)+5(x-7)$$ = (x - 7)(x + 5)$Hence, $7$ and $-5 $are the other zeros of this polynomial.

Question 6

Obtain all other zeroes of $3x^4 +6x^3-2x^2-10x-5$, if two of its zeroes are $\sqrt{\dfrac{5}{3}}$ and $-\sqrt{\dfrac{5}{3}}$.

Accepted Answer

Two zeroes are $\sqrt{\dfrac{5}{3}}$ and $-\sqrt{\dfrac{5}{3}}$So we can write it as, x = $\sqrt{\dfrac{5}{3}}$ and x = $-\sqrt{\dfrac{5}{3}}$we get $x-\sqrt{\dfrac{5}{3}}=0$ and $x+\sqrt{\dfrac{5}{3}}=0$Multiply both the factors we get,$x^2-\dfrac{5}{3}=0$Multiply by $3$ we get$3x^2-5 = 0$ is the factor of $3x^4 +6x^3-2x^2-10x-5$Now divide, $3x^4 +6x^3-2x^2-10x-5$ by $3x^2-5 = 0$ we get,Quotient is $x^2+2x+1=0$Compare the equation with quadratic formula,$x^2 - (Sum \space\ of \space\ root)x + (Product \space\ of \space\ root) = 0$$\Rightarrow$Sum of root $= 2$$\Rightarrow$Product of the root $= 1$So, we get$\Rightarrow$$x^2+ x+x+1 = 0$$\Rightarrow$$x(x + 1) + 1(x + 1) = 0$$\Rightarrow$$x + 1 = 0, x + 1= 0$$\Rightarrow$$x = -1, x = -1$So, our zeroes are $-1, -1,$ $\sqrt{\dfrac{5}{3}}$ and $-\sqrt{\dfrac{5}{3}}$

Question 7

Expand the following:$(4a - b + 2c)^2$

Accepted Answer

We have,$(4a - b + 2c)^2 \= (4a)^2 + (-b)^2 + (2c)^2 + 2(4a)(-b) + 2(-b)(2c) + 2(2c)(4a)$$\qquad[\because a^2 + b^2 + c^2 + 2ab + 2bc + 2ca = (a + b + c)^2 ]$$= 16a^2 + b^2 + 4c^2 - 8ab - 4ac + 16ca$

Question 8

Simplify:$(7m -8n)^2 + (7m + 8n)^2$.

Accepted Answer

Given, $(7m - 8n)^2 + (7m + 8n)^2$.We know, $(a-b)^2=a^2-2ab+b^2$ and $(a+b)^2=a^2+2ab+b^2$.Then,$(7m - 8n)^2 + (7m + 8n)^2$$=[(7m)^2-2(7m)(8n)+(8n)^2]+[(7m)^2+2(7m)(8n)+(8n)^2]$$=(49m^2-112mn+64n^2)+(49m^2+112mn+64n^2)$$=49m^2-112mn+64n^2+49m^2+112mn+64n^2$$=98m^2+128n^2$.Therefore, option $B$ is correct.

Question 9

Show that: $(4pq + 3q)^2 -(4pq -3q)^2 = 48pq^2$

Accepted Answer

$(4pq + 3q)^2 -(4pq -3q)^2 = 48pq^2$LHS$=(4pq)^2+2(4pq)(3q)+(3q)^2 - [(4pq)^2-2(4pq)(3q)+(3q)^2]$       $=16p^{ 2 }q^{ 2 }+24pq^{ 2 }+9q^2-16p^2q^2+24pq^2-9q^2$       $=48pq^2$           =RHSLHS=RHS

Question 10

Show that: $(9p - 5q)^2 + 180pq = (9p + 5q)^2$

Accepted Answer

$(9p - 5q)^2 + 180pq = (9p + 5q)^2$LHS$=(9p)^2-2(9p)(5q)+(5q)^2+180pq$       $= 81p^2 - 90pq + 25q^2 + 180pq$       $= 81p^2 + 90pq + 25q^2$RHS$= (9p + 5q)^2$       $=(9p)^2+2(9p)(5q)+(5q)^2$       $= 81p^2 + 90pq + 25q^2$LHS=RHS