Polynomials
Category : 10th Class
Polynomials
Polynomials
e.g.\[{{x}^{2}}-5,\,\,\,5\sqrt{2}{{x}^{2}}-\frac{1}{\sqrt{3}}x,\,\,7{{x}^{2}}\,\,+\,\,\sqrt{5}\] etc. are quadratic polynomials.
e.g. \[{{x}^{3}}-20,\,\,\sqrt{5}{{x}^{3}}\,\,-\,\,\frac{1}{9}x,\,\,\frac{7}{2}{{x}^{3}}-\frac{1}{2}{{x}^{2}}-4\]etc. are cubic polynomials.
e.g. \[{{x}^{4}}\text{ }23,\,\sqrt{3}{{x}^{4}}-\frac{1}{9}x,\,\,\frac{1}{2}{{x}^{4}}+\frac{3}{4}x-\frac{1}{8}\] etc. are biquadratic polynomials.
\[p\left( x \right)\text{ }=\text{ }ax\text{ }+\text{ }b\] is given by\[\alpha =\frac{-b}{a}=\frac{-(constant\,\,term)}{(coefficient\,\,of\,\,x)}\].
A linear polynomial can have at the most one zero.
(i) If \[\alpha \] and \[\beta \] are the zeros of a quadratic polynomial \[p\left( x \right)\text{ }=\text{ }a{{x}^{2}}+\text{ }bx\text{ }+\text{ }c,\text{ }a\,\,\ne \,~0\] then
\[\alpha +\beta \,\,=\,\,\frac{-b}{a}\,\,=\,\,\frac{-(coefficient\,\,of\,\,x)}{(coefficien\,\,of\,\,{{x}^{2}})};\]
(ii) A quadratic polynomial whose zeroes are \[\alpha \] and \[\beta \] is given by:
\[p\left( x \right)\text{ }=\text{ }{{x}^{2}}\,-\,\,(\alpha +\beta )\,\,+\text{ }(\alpha \beta )\]
(i) If \[\alpha ,\,\,\beta \,\,and\,\,\gamma \] are the zeros of\[p\left( x \right)\text{ }=\text{ }a{{x}^{3}}+\text{ }b{{x}^{2}}+\text{ }cx\text{ }+\text{ }d\], then
\[(\alpha \,\,+\,\,\beta \,\,+\,\,\gamma )\,\,=\,\,\frac{-b}{a}\,\,=\,\,\frac{-(coefficient\,\,of\,\,{{x}^{2}})}{(coefficient\,\,of\,\,{{x}^{3}})};\]
\[(\alpha \beta \,\,+\,\,\beta \gamma \,\,+\,\,\gamma \alpha )\,\,=\,\,\frac{c}{a}\,\,=\,\,\frac{(coefficient\,\,of\,\,x)}{(coefficient\,\,of\,\,{{x}^{3}})};\]
\[\alpha \beta \gamma \,\,=\,\,\frac{-d}{a}\,\,=\,\,\frac{-(cons\tan t\,\,term)}{(coefficient\,\,of\,\,{{x}^{3}})}\]
(ii) A cubic polynomial whose zeros are \[\alpha ,\,\,\beta \,\,and\,\,\gamma \] is given by
\[p(x)=\{{{x}^{3}}-(\alpha +\beta +\gamma ){{x}^{2}}+(\alpha \beta +\beta \gamma +\gamma \alpha )x-\alpha \beta \gamma \}\]
Snap Test
(a) \[f\left( 0 \right)\text{ }=\text{ }a\]
(b) \[f\left( a \right)\text{ }-\text{ }a\]
(c)\[f\left( a \right)=0\]
(d) \[f\left( a \right)=f\left( 0 \right)\]
(e) None of these
Ans. (c)
Explanation: We know that for the polynomial f(x), if f\[\left( a \right)\text{ }=\text{ }0\], then a is a zero of the polynomial f(x)
(a) \[{{x}^{2}}+\text{ }6x\text{ }+\text{ }6\]
(b) \[{{x}^{2}}-\text{ }6x\text{ }+\text{ }6\]
(c) \[{{x}^{2}}+\text{ }2x\text{ }+\text{ }4\]
(d) \[x\text{ }+\text{ }6x\text{ }-\text{ }6\]
(e) None of these
Ans. (a)
Explanation: Required polynomial
\[f(x)\,\,\,\,=\,\,\,\,[x-(-\,3\,\,+\,\,\sqrt{3})]\,\,[x-(-\,\,3\,\,-\sqrt{3})]\]
= \[[(x+3)-\sqrt{3}]\,\,\,[(x+3)+\sqrt{3}]\]
= \[{{(x+3)}^{2}}-{{(\sqrt{3})}^{2}}\] \[[\,\,\because \,\,\,~\left( a\text{ }-\text{ }b \right)\left( a\text{ }+\text{ }b \right)\text{ }=\text{ }{{a}^{2}}-\text{ }{{b}^{2}}]\]
\[\,=\,\,\,\,\,\left( {{x}^{2}}+\text{ }6x\text{ }+\text{ }9 \right)\text{ }-\text{ }3\text{ }=\text{ }{{x}^{2}}+\text{ }6x\text{ }+\text{ }6\].
(a) \[\frac{-10}{35}\] (b) \[\frac{-12}{35}\]
(c) \[\frac{-14}{35}\] (d) \[\frac{-11}{35}\]
(e) None of these
Ans. (b)
Explanation: Given that \[~\alpha \,and\,\beta \] are the zeros of the polynomial \[p\left( x \right)=\text{ }{{x}^{2}}+\text{ }12x\text{ }+\text{ }35\], Therefore, \[~\alpha +\beta \,\,~=-\text{ }12\] and\[~\alpha \beta \,\,=\text{ }35\].
\[\therefore \,\,\,\frac{1}{\alpha }\,\,+\,\,\frac{1}{\beta }\,\,=\,\,\frac{\alpha +\beta }{\alpha \beta }\,\,=\,\,\frac{-12}{35}\]
(a) \[k\text{ }=\text{ }0\]
(b) \[k\text{ }=\text{ }1\]
(c) \[k\text{ }=\text{ }3\]
(d) \[k\text{ }=\text{ }2\]
(e) None of these
Ans. (a)
Explanation: Let \[~\alpha \,\,and\,\,\beta \] be the zeros of the polynomial, \[f\left( x \right)=\text{ }14{{x}^{2}}-\text{ }42\text{ }{{k}^{2}}x\text{ }-\text{ }9\].
Then \[\alpha +\beta \,\,\,\,\,\,\,=\,\,\,\,\,\frac{42{{k}^{2}}}{14}\,\,\,\,=\,\,\,\,\,\,3{{k}^{2}}\] [Sum of the zeros of f(x)]
Now, let \[\beta \,\,\,\,\,\,\,\,=\,\,\,\,\,\,\,(-\alpha )\] [Since one of the zeros of f(x) is negative of the other]
Then, \[~~\alpha \,\,+\,\,\beta ~\,\,=\text{ }0\].
Equating the two values of \[(\alpha +\beta )\] we get:
\[3{{k}^{2}}=\text{ }0\text{ }\Rightarrow \text{ }{{k}^{2}}=\text{ }0\text{ }\Rightarrow \text{ }k\text{ }=\text{ }0\]
(a) -2, 3 and 7 (b) -1, 2 and 5
(c) -3, 2 and 7 (d) -4, 2 and 5
(e) None of these
Ans. (c)
Explanation: Let \[(\alpha -d),\alpha \,\,and\,(\alpha +d)\] be the zeros of the polynomial
\[f\left( x \right)\text{ }=\text{ }{{x}^{3}}-\text{ }6{{x}^{2}}-\text{ }13x\text{ }+\text{ }42\] (Since the zeros are in A.P.)
Then, Sum of the zeros of \[f\left( x \right)\text{ }=\text{ }6\]
i.e. \[\left( \alpha \text{ }-\text{ }d \right)\text{ }+\text{ }\alpha \text{ }+\text{ }\left( \alpha \text{ }+\text{ }d \right)\text{ }=\text{ }6\text{ }\Rightarrow \text{ }3\alpha \text{ }=\,\,\,6\,\,\,\Rightarrow \text{ }\alpha \,\,=\text{ }2\]
Also, Product of the zeros of \[f\left( x \right)\text{ }=\text{ }-\text{ }42\]
i.e. \[\left( \alpha \text{ }-\text{ }d \right)\text{ }\alpha \left( \alpha \text{ }+\text{ }d \right)\text{ }=\text{ }-\text{ }42\]
\[\Rightarrow \,\,\,\,\alpha \left( {{\alpha }^{2}}-{{d}^{2}} \right)=-\,42\,\,\Rightarrow \,\,2\left( {{2}^{2}}-{{d}^{2}} \right)\,\,=-\,42\]
\[\Rightarrow \,\,{{d}^{2}}=\text{ }25\text{ }\Rightarrow \text{ }d\,\,=\pm \text{ }5\]
Taking any of the values of d i.e. taking either \[d\,\,=\,\,5\] or \[d\text{ }=\text{ }-\text{ }5\], we get the zeros of f (x) as - 3, 2 and 7.
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