\displaystyle \textbf{Question 1. }\text{Check whether the first polynomial is a factor of the second polynomial} \\ \text{by applying the division algorithm.}

\displaystyle \textbf{(i) }g(t)=t^2-3,\ f(t)=2t^4+3t^3-2t^2-9t-12
\displaystyle \text{Answer:}
\displaystyle \text{Dividing }f(t)\text{ by }g(t),\text{ we get}
\displaystyle f(t)=(t^2-3)(2t^2+3t+4)+0
\displaystyle \therefore \text{Remainder}=0
\displaystyle \text{Hence, }t^2-3\text{ is a factor of }2t^4+3t^3-2t^2-9t-12.
\\

\displaystyle \textbf{(ii) }g(x)=x^3-3x+1,\ f(x)=x^5-4x^3+x^2+3x+1
\displaystyle \text{Answer:}
\displaystyle \text{Dividing }f(x)\text{ by }g(x),\text{ we get}
\displaystyle f(x)=(x^3-3x+1)(x^2-1)+2
\displaystyle \therefore \text{Remainder}=2
\displaystyle \text{Since the remainder is not zero, }x^3-3x+1\text{ is not a factor of }x^5-4x^3+x^2+3x+1.
\\

\displaystyle \textbf{(iii) }g(x)=2x^2-x+3,\ f(x)=6x^5-x^4+4x^3-5x^2-x-15\qquad [\mathrm{CBSE}\ 2018]
\displaystyle \text{Answer:}
\displaystyle \text{Dividing }f(x)\text{ by }g(x),\text{ we get}
\displaystyle f(x)=(2x^2-x+3)(3x^3+x^2-2x-5)+0
\displaystyle \therefore \text{Remainder}=0
\displaystyle \text{Hence, }2x^2-x+3\text{ is a factor of }6x^5-x^4+4x^3-5x^2-x-15.
\\

\displaystyle \textbf{Question 2. }\text{Obtain all zeros of }f(x)=x^3+13x^2+32x+20, \\ \text{if one of its zeros is }-2.
\displaystyle \text{Answer:}
\displaystyle \text{Since }-2\text{ is a zero, }x+2\text{ is a factor of }f(x).
\displaystyle f(x)=x^3+13x^2+32x+20
\displaystyle =(x+2)(x^2+11x+10)
\displaystyle =(x+2)(x^2+10x+x+10)
\displaystyle =(x+2)(x+10)(x+1)
\displaystyle \text{Hence, all zeros are }-2,-10\text{ and }-1.
\\

\displaystyle \textbf{Question 3. }\text{Obtain all zeros of }f(x)=2x^4+x^3-14x^2-19x-6,\text{ if two of its zeros are }
\displaystyle -2 \ \text{and }-1.
\displaystyle \text{Answer:}
\displaystyle \text{Since }-2\text{ and }-1\text{ are zeros, }(x+2)(x+1)\text{ is a factor of }f(x).
\displaystyle (x+2)(x+1)=x^2+3x+2
\displaystyle f(x)=2x^4+x^3-14x^2-19x-6
\displaystyle =(x^2+3x+2)(2x^2-5x-3)
\displaystyle =(x+2)(x+1)(2x^2-6x+x-3)
\displaystyle =(x+2)(x+1)\{2x(x-3)+1(x-3)\}
\displaystyle =(x+2)(x+1)(x-3)(2x+1)
\displaystyle \text{Hence, all zeros are }-2,-1,3\text{ and }-\frac{1}{2}.
\\

\displaystyle \textbf{Question 4. }\text{Find all the zeros of the polynomial }x^4+x^3-34x^2-4x+120,
\displaystyle \text{ if two of its zeros are }2 \text{ and }-2.\qquad [\mathrm{CBSE}\ 2008]
\displaystyle \text{Answer:}
\displaystyle \text{Since }2\text{ and }-2\text{ are zeros, }(x-2)(x+2)\text{ is a factor of the given polynomial.}
\displaystyle (x-2)(x+2)=x^2-4
\displaystyle x^4+x^3-34x^2-4x+120
\displaystyle =(x^2-4)(x^2+x-30)
\displaystyle =(x-2)(x+2)(x^2+6x-5x-30)
\displaystyle =(x-2)(x+2)\{x(x+6)-5(x+6)\}
\displaystyle =(x-2)(x+2)(x+6)(x-5)
\displaystyle \text{Hence, all zeros are }2,-2,-6\text{ and }5.
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\displaystyle \textbf{Question 5. }\text{For what value of }k,\text{ is the polynomial }
\displaystyle f(x)=3x^4-9x^3+x^2+15x+k\text{ completely divisible by } 3x^2-5?\qquad [\mathrm{CBSE}\ 2019]
\displaystyle \text{Answer:}
\displaystyle \text{Since }f(x)\text{ is completely divisible by }3x^2-5,\text{ remainder}=0.
\displaystyle \text{Dividing }f(x)\text{ by }3x^2-5,\text{ the remainder is }k+10.
\displaystyle \therefore k+10=0
\displaystyle \Rightarrow k=-10
\displaystyle \text{Hence, }k=-10.
\\

\displaystyle \textbf{Question 6. }\text{Find all the zeros of the polynomial }2x^3+x^2-6x-3,
\displaystyle \text{ if two of its zeros are }-\sqrt{3} \ \text{and }\sqrt{3}.\qquad [\mathrm{CBSE}\ 2009]
\displaystyle \text{Answer:}
\displaystyle \text{Since }-\sqrt{3}\text{ and }\sqrt{3}\text{ are zeros, }(x+\sqrt{3})(x-\sqrt{3})\text{ is a factor.}
\displaystyle (x+\sqrt{3})(x-\sqrt{3})=x^2-3
\displaystyle 2x^3+x^2-6x-3
\displaystyle =(x^2-3)(2x+1)
\displaystyle =(x+\sqrt{3})(x-\sqrt{3})(2x+1)
\displaystyle \text{The zeros are given by }f(x)=0.
\displaystyle (x+\sqrt{3})(x-\sqrt{3})(2x+1)=0
\displaystyle \Rightarrow x=-\sqrt{3},\sqrt{3},-\frac{1}{2}
\displaystyle \text{Hence, all zeros are }-\sqrt{3},\sqrt{3}\text{ and }-\frac{1}{2}.
\\

\displaystyle \textbf{Question 7. }\text{Find all the zeros of the polynomial }x^3+3x^2-2x-6,
\displaystyle \text{ if two of its zeros are }-\sqrt{2} \text{ and }\sqrt{2}.\qquad [\mathrm{CBSE}\ 2009]
\displaystyle \text{Answer:}
\displaystyle \text{Since }-\sqrt{2}\text{ and }\sqrt{2}\text{ are zeros, }(x+\sqrt{2})(x-\sqrt{2})\text{ is a factor.}
\displaystyle (x+\sqrt{2})(x-\sqrt{2})=x^2-2
\displaystyle x^3+3x^2-2x-6
\displaystyle =(x^2-2)(x+3)
\displaystyle =(x+\sqrt{2})(x-\sqrt{2})(x+3)
\displaystyle \text{The zeros are given by }f(x)=0.
\displaystyle (x+\sqrt{2})(x-\sqrt{2})(x+3)=0
\displaystyle \Rightarrow x=-\sqrt{2},\sqrt{2},-3
\displaystyle \text{Hence, all zeros are }-\sqrt{2},\sqrt{2}\text{ and }-3.
\\

\displaystyle \textbf{Question 8. }\text{Find all zeros of the polynomial }
\displaystyle f(x)=2x^4-2x^3-7x^2+3x+6,\text{ if its two zeros are }-\sqrt{\frac{3}{2}} \text{ and }\sqrt{\frac{3}{2}}.
\displaystyle \text{Answer:}
\displaystyle \text{Since }-\sqrt{\frac{3}{2}}\text{ and }\sqrt{\frac{3}{2}}\text{ are zeros, }x^2-\frac{3}{2}\text{ is a factor.}
\displaystyle x^2-\frac{3}{2}=\frac{1}{2}(2x^2-3)
\displaystyle \therefore 2x^2-3\text{ is a factor of }f(x).
\displaystyle 2x^4-2x^3-7x^2+3x+6
\displaystyle =(2x^2-3)(x^2-x-2)
\displaystyle =(2x^2-3)(x-2)(x+1)
\displaystyle \text{The zeros are given by }f(x)=0.
\displaystyle (2x^2-3)(x-2)(x+1)=0
\displaystyle \Rightarrow x=-\sqrt{\frac{3}{2}},\sqrt{\frac{3}{2}},2,-1
\displaystyle \text{Hence, all zeros are }-\sqrt{\frac{3}{2}},\sqrt{\frac{3}{2}},2\text{ and }-1.
\\

\displaystyle \textbf{Question 9. }\text{Find all zeros of the polynomial }2x^4+7x^3-19x^2-14x+30,
\displaystyle \text{ if two of its zeros are }\sqrt{2} \text{ and }-\sqrt{2}.\qquad [\mathrm{CBSE}\ 2008]
\displaystyle \text{Answer:}
\displaystyle \text{Since }\sqrt{2}\text{ and }-\sqrt{2}\text{ are zeros, }(x-\sqrt{2})(x+\sqrt{2})\text{ is a factor.}
\displaystyle (x-\sqrt{2})(x+\sqrt{2})=x^2-2
\displaystyle 2x^4+7x^3-19x^2-14x+30
\displaystyle =(x^2-2)(2x^2+7x-15)
\displaystyle =(x^2-2)(2x-3)(x+5)
\displaystyle \text{The zeros are given by }f(x)=0.
\displaystyle (x-\sqrt{2})(x+\sqrt{2})(2x-3)(x+5)=0
\displaystyle \Rightarrow x=\sqrt{2},-\sqrt{2},\frac{3}{2},-5
\displaystyle \text{Hence, all zeros are }\sqrt{2},-\sqrt{2},\frac{3}{2}\text{ and }-5.
\\

\displaystyle \textbf{Question 10. }\text{Given that }x-\sqrt{5}\text{ is a factor of the cubic polynomial }
\displaystyle x^3-3\sqrt{5}x^2+13x-3\sqrt{5}, \ \text{find all the zeroes of the polynomial.}
\displaystyle \text{Answer:}
\displaystyle \text{Since }x-\sqrt{5}\text{ is a factor, }\sqrt{5}\text{ is a zero of the polynomial.}
\displaystyle x^3-3\sqrt{5}x^2+13x-3\sqrt{5}
\displaystyle =(x-\sqrt{5})(x^2-2\sqrt{5}x+3)
\displaystyle \text{Now,}
\displaystyle x^2-2\sqrt{5}x+3=0
\displaystyle \Rightarrow x=\frac{2\sqrt{5}\pm\sqrt{(2\sqrt{5})^2-4(1)(3)}}{2}
\displaystyle =\frac{2\sqrt{5}\pm\sqrt{20-12}}{2}
\displaystyle =\frac{2\sqrt{5}\pm2\sqrt{2}}{2}
\displaystyle =\sqrt{5}\pm\sqrt{2}
\displaystyle \text{Hence, all the zeroes are }\sqrt{5},\sqrt{5}+\sqrt{2}\text{ and }\sqrt{5}-\sqrt{2}.
\\

\displaystyle \textbf{Question 11. }\text{What must be added to the polynomial }f(x)=x^4+2x^3-2x^2+x-1\text{ so that the resulting}
\displaystyle \text{polynomial is exactly divisible by }x^2+2x-3?

\displaystyle \text{Answer:}
\displaystyle \text{Let }g(x)=x^2+2x-3.
\displaystyle \text{On dividing }f(x)\text{ by }g(x)\text{ using long division, we obtain}
\displaystyle f(x)=(x^2+2x-3)(x^2+1)+(2-x)
\displaystyle \therefore \text{Quotient}=x^2+1,\qquad \text{Remainder}=2-x.
\displaystyle \text{For exact divisibility, the remainder must be zero.}
\displaystyle \text{By division algorithm,}
\displaystyle \text{Dividend}=\text{Divisor}\times\text{Quotient}+\text{Remainder}
\displaystyle \text{Clearly, RHS is divisible by the divisor. Therefore, if we subtract the remainder}
\displaystyle \text{from the dividend, the resulting polynomial will also be divisible by the divisor.}
\displaystyle \text{Hence, we must add }-r(x)\text{ to }f(x).
\displaystyle -r(x)=-(2-x)=x-2
\displaystyle \therefore \boxed{x-2}\text{ must be added to the polynomial.}
\\

\displaystyle \textbf{Question 12. }\text{Apply division algorithm to find the quotient }q(x) \\ \text{and remainder }r(x)\text{ on dividing }f(x)\text{ by }g(x).

\displaystyle \textbf{(i) }f(x)=x^3-6x^2+11x-6,\quad g(x)=x^2+x+1
\displaystyle \text{Answer:}
\displaystyle \text{Using division algorithm, we obtain}
\displaystyle f(x)=g(x)\,q(x)+r(x)
\displaystyle x^3-6x^2+11x-6=(x^2+x+1)(x-7)+(17x+1)
\displaystyle \therefore q(x)=x-7,\qquad r(x)=17x+1.
\\

\displaystyle \textbf{(ii) }f(x)=10x^4+17x^3-62x^2+30x-3,\quad g(x)=2x^2+7x+1
\displaystyle \text{Answer:}
\displaystyle \text{Using division algorithm, we obtain}
\displaystyle f(x)=g(x)\,q(x)+r(x)
\displaystyle 10x^4+17x^3-62x^2+30x-3=(2x^2+7x+1)(5x^2-9x-2)+(53x-1)
\displaystyle \therefore q(x)=5x^2-9x-2,\qquad r(x)=53x-1.
\\

\displaystyle \textbf{(iii) }f(x)=4x^3+8x^2+8x+7,\quad g(x)=2x^2-x+1
\displaystyle \text{Answer:}
\displaystyle \text{Using division algorithm, we obtain}
\displaystyle f(x)=g(x)\,q(x)+r(x)
\displaystyle 4x^3+8x^2+8x+7=(2x^2-x+1)(2x+5)+(11x+2)
\displaystyle \therefore q(x)=2x+5,\qquad r(x)=11x+2.
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\displaystyle \textbf{(iv) }f(x)=15x^3-20x^2+13x-12,\quad g(x)=x^2-2x+2
\displaystyle \text{Answer:}
\displaystyle \text{Using division algorithm, we obtain}
\displaystyle f(x)=g(x)\,q(x)+r(x)
\displaystyle 15x^3-20x^2+13x-12=(x^2-2x+2)(15x+10)+(3x-32)
\displaystyle \therefore q(x)=15x+10,\qquad r(x)=3x-32.
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