Polynomials

$\displaystyle \textbf{Question 1. }\text{Verify that the numbers given along side of the cubic polynomials below }$
$\displaystyle \text{are their zeros. Also, verify the relationship between the zeros and coefficients in each case.}$

$\displaystyle \textbf{(i) }f(x)=2x^3+x^2-5x+2;\ \frac{1}{2},1,-2$
$\displaystyle \text{Answer:}$
$\displaystyle f(x)=2x^3+x^2-5x+2$
$\displaystyle f\left(\frac{1}{2}\right)=2\left(\frac{1}{2}\right)^3+\left(\frac{1}{2}\right)^2-5\left(\frac{1}{2}\right)+2$
$\displaystyle =\frac{1}{4}+\frac{1}{4}-\frac{5}{2}+2=0$
$\displaystyle f(1)=2(1)^3+(1)^2-5(1)+2=2+1-5+2=0$
$\displaystyle f(-2)=2(-2)^3+(-2)^2-5(-2)+2=-16+4+10+2=0$
$\displaystyle \text{So, }\frac{1}{2},1,-2\text{ are the zeros of }f(x).$
$\displaystyle \text{Let }\alpha=\frac{1}{2},\ \beta=1,\ \gamma=-2.$
$\displaystyle \alpha+\beta+\gamma=\frac{1}{2}+1-2=-\frac{1}{2}$
$\displaystyle \alpha\beta+\beta\gamma+\gamma\alpha=\frac{1}{2}(1)+1(-2)+(-2)\left(\frac{1}{2}\right)$
$\displaystyle =\frac{1}{2}-2-1=-\frac{5}{2}$
$\displaystyle \alpha\beta\gamma=\frac{1}{2}\times1\times(-2)=-1\qquad ...(i)$
$\displaystyle \text{From }f(x)=2x^3+x^2-5x+2,\text{ we find that}$
$\displaystyle \text{Coefficient of }x^3=2,\text{ Coefficient of }x^2=1,\text{ Coefficient of }x=-5\text{ and Constant term}=2$
$\displaystyle -\frac{\text{Coefficient of }x^2}{\text{Coefficient of }x^3}=-\frac{1}{2}$
$\displaystyle \frac{\text{Coefficient of }x}{\text{Coefficient of }x^3}=\frac{-5}{2}=-\frac{5}{2}$
$\displaystyle -\frac{\text{Constant term}}{\text{Coefficient of }x^3}=-\frac{2}{2}=-1\qquad ...(ii)$
$\displaystyle \text{From (i) and (ii), the relationship between zeros and coefficients is verified.}$
$\\$

$\displaystyle \textbf{(ii) }g(x)=x^3-4x^2+5x-2;\ 2,1,1$
$\displaystyle \text{Answer:}$
$\displaystyle g(x)=x^3-4x^2+5x-2$
$\displaystyle g(2)=2^3-4(2)^2+5(2)-2=8-16+10-2=0$
$\displaystyle g(1)=1^3-4(1)^2+5(1)-2=1-4+5-2=0$
$\displaystyle \text{So, }2,1,1\text{ are the zeros of }g(x).$
$\displaystyle \text{Let }\alpha=2,\ \beta=1,\ \gamma=1.$
$\displaystyle \alpha+\beta+\gamma=2+1+1=4$
$\displaystyle \alpha\beta+\beta\gamma+\gamma\alpha=2(1)+1(1)+1(2)=5$
$\displaystyle \alpha\beta\gamma=2\times1\times1=2\qquad ...(i)$
$\displaystyle \text{From }g(x)=x^3-4x^2+5x-2,\text{ we find that}$
$\displaystyle \text{Coefficient of }x^3=1,\text{ Coefficient of }x^2=-4,\text{ Coefficient of }x=5\text{ and Constant term}=-2$
$\displaystyle -\frac{\text{Coefficient of }x^2}{\text{Coefficient of }x^3}=-\frac{-4}{1}=4$
$\displaystyle \frac{\text{Coefficient of }x}{\text{Coefficient of }x^3}=\frac{5}{1}=5$
$\displaystyle -\frac{\text{Constant term}}{\text{Coefficient of }x^3}=-\frac{-2}{1}=2\qquad ...(ii)$
$\displaystyle \text{From (i) and (ii), the relationship between zeros and coefficients is verified.}$
$\\$

$\displaystyle \textbf{Question 2. }\text{Find a cubic polynomial with the sum, sum of the product of its zeros}$
$\displaystyle \text{taken two at a time, and product of its zeros as }3,-1\text{ and }-3\text{ respectively.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{If }\alpha,\beta,\gamma\text{ are the zeros of a cubic polynomial }f(x),\text{ then}$
$\displaystyle f(x)=k\{x^3-(\alpha+\beta+\gamma)x^2+(\alpha\beta+\beta\gamma+\gamma\alpha)x-\alpha\beta\gamma\}$
$\displaystyle \text{Here, }\alpha+\beta+\gamma=3,\quad \alpha\beta+\beta\gamma+\gamma\alpha=-1,\quad \alpha\beta\gamma=-3$
$\displaystyle \text{Substituting these values, we get}$
$\displaystyle f(x)=k\{x^3-3x^2-x+3\}$
$\displaystyle \text{where }k\text{ is any non-zero real number.}$
$\displaystyle \text{Taking }k=1,\text{ the required cubic polynomial is}$
$\displaystyle x^3-3x^2-x+3.$
$\\$

$\displaystyle \textbf{Question 3. }\text{If the zeroes of the polynomial } \\ f(x)=2x^3-15x^2+37x-30\text{ are in A.P., find them.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }\alpha=a-d,\ \beta=a,\ \gamma=a+d\text{ be the zeroes of }f(x).$
$\displaystyle \alpha+\beta+\gamma=-\frac{-15}{2}=\frac{15}{2}$
$\displaystyle \Rightarrow (a-d)+a+(a+d)=\frac{15}{2}$
$\displaystyle \Rightarrow 3a=\frac{15}{2}$
$\displaystyle \Rightarrow a=\frac{5}{2}$
$\displaystyle \alpha\beta\gamma=-\frac{-30}{2}=15$
$\displaystyle \Rightarrow (a-d)a(a+d)=15$
$\displaystyle \Rightarrow a(a^2-d^2)=15$
$\displaystyle \Rightarrow \frac{5}{2}\left(\frac{25}{4}-d^2\right)=15$
$\displaystyle \Rightarrow \frac{25}{4}-d^2=6$
$\displaystyle \Rightarrow d^2=\frac{1}{4}$
$\displaystyle \Rightarrow d=\pm\frac{1}{2}$
$\displaystyle \text{Hence, the zeroes are }2,\frac{5}{2},3.$
$\\$

$\displaystyle \textbf{Question 4. }\text{Find the condition that the zeroes of the polynomial } \\ f(x)=x^3+3px^2+3qx+r\text{ may be in A.P.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }a-d,\ a,\ a+d\text{ be the zeroes of }f(x).$
$\displaystyle \text{Sum of zeroes}=-\frac{3p}{1}=-3p$
$\displaystyle \Rightarrow (a-d)+a+(a+d)=-3p$
$\displaystyle \Rightarrow 3a=-3p$
$\displaystyle \Rightarrow a=-p$
$\displaystyle \text{Since }a\text{ is a zero of }f(x),\text{ therefore }f(a)=0.$
$\displaystyle \Rightarrow f(-p)=0$
$\displaystyle \Rightarrow (-p)^3+3p(-p)^2+3q(-p)+r=0$
$\displaystyle \Rightarrow -p^3+3p^3-3pq+r=0$
$\displaystyle \Rightarrow 2p^3-3pq+r=0$
$\displaystyle \text{Hence, the required condition is }2p^3-3pq+r=0.$
$\\$

$\displaystyle \textbf{Question 5. }\text{If the zeroes of the polynomial }f(x)=ax^3+3bx^2+3cx+d$
$\displaystyle \text{ are in A.P., prove that } 2b^3-3abc+a^2d=0.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }m-d,\ m,\ m+d\text{ be the zeroes of }f(x).$
$\displaystyle \text{Sum of zeroes}=-\frac{3b}{a}$
$\displaystyle \Rightarrow (m-d)+m+(m+d)=-\frac{3b}{a}$
$\displaystyle \Rightarrow 3m=-\frac{3b}{a}$
$\displaystyle \Rightarrow m=-\frac{b}{a}$
$\displaystyle \text{Since }m\text{ is a zero of }f(x),\text{ therefore }f(m)=0.$
$\displaystyle \Rightarrow f\left(-\frac{b}{a}\right)=0$
$\displaystyle \Rightarrow a\left(-\frac{b}{a}\right)^3+3b\left(-\frac{b}{a}\right)^2+3c\left(-\frac{b}{a}\right)+d=0$
$\displaystyle \Rightarrow -\frac{b^3}{a^2}+\frac{3b^3}{a^2}-\frac{3bc}{a}+d=0$
$\displaystyle \Rightarrow \frac{2b^3}{a^2}-\frac{3bc}{a}+d=0$
$\displaystyle \Rightarrow 2b^3-3abc+a^2d=0$
$\displaystyle \text{Hence proved.}$
$\\$

$\displaystyle \textbf{Question 6. }\text{If the zeroes of the polynomial }f(x)=x^3-12x^2+39x+k \\ \text{are in A.P., find the value of }k.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }\alpha=a-d,\ \beta=a,\ \gamma=a+d\text{ be the zeroes of }f(x).$
$\displaystyle \alpha+\beta+\gamma=-\frac{-12}{1}=12$
$\displaystyle \Rightarrow (a-d)+a+(a+d)=12$
$\displaystyle \Rightarrow 3a=12$
$\displaystyle \Rightarrow a=4$
$\displaystyle \alpha\beta+\beta\gamma+\gamma\alpha=\frac{39}{1}=39$
$\displaystyle \Rightarrow (a-d)a+a(a+d)+(a-d)(a+d)=39$
$\displaystyle \Rightarrow 3a^2-d^2=39$
$\displaystyle \Rightarrow 3(4)^2-d^2=39$
$\displaystyle \Rightarrow 48-d^2=39$
$\displaystyle \Rightarrow d^2=9$
$\displaystyle \Rightarrow d=\pm3$
$\displaystyle \therefore \alpha\beta\gamma=(a-d)a(a+d)=4(16-9)=28$
$\displaystyle \text{Also, }\alpha\beta\gamma=-\frac{k}{1}=-k$
$\displaystyle \Rightarrow -k=28$
$\displaystyle \Rightarrow k=-28$
$\displaystyle \text{Hence, }k=-28.$
$\\$

$\displaystyle \textbf{Question 7. }\text{If }4\text{ is a zero of the cubic polynomial }x^3-3x^2-10x+24,\text{ find its other two zeroes.}$
$\displaystyle \qquad [\mathrm{CBSE}\ 2020]$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }\alpha,\ \beta\text{ be the other two zeroes of the given polynomial. Then,}$
$\displaystyle \alpha+\beta+4=-\frac{-3}{1}=3\qquad ...(i)$
$\displaystyle \alpha\beta+4\alpha+4\beta=\frac{-10}{1}=-10\qquad ...(ii)$
$\displaystyle \alpha\beta(4)=-\frac{24}{1}=-24\qquad ...(iii)$
$\displaystyle \text{From (iii),}$
$\displaystyle 4\alpha\beta=-24$
$\displaystyle \Rightarrow \alpha\beta=-6$
$\displaystyle \text{From (i),}$
$\displaystyle \alpha+\beta+4=3$
$\displaystyle \Rightarrow \alpha+\beta=-1$
$\displaystyle \text{A quadratic polynomial having }\alpha,\beta\text{ as its zeroes is}$
$\displaystyle f(x)=x^2-(\alpha+\beta)x+\alpha\beta$
$\displaystyle =x^2-(-1)x-6$
$\displaystyle =x^2+x-6$
$\displaystyle =x^2+3x-2x-6$
$\displaystyle =x(x+3)-2(x+3)$
$\displaystyle =(x+3)(x-2)$
$\displaystyle \text{Hence, the other two zeroes are }-3\text{ and }2.$
$\\$