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Home » Find the zeroes of the following polynomials by factorisation method. 2×2 +(7/2)x +3/4
Find the zeroes of the following polynomials by factorisation method. 2×2 +(7/2)x +3/4
CBSE | Class 10 | Exercise 2.3 | Maths | NCERT Exemplar | Polynomials
Find the zeroes of the following polynomials by factorisation method. 2×2 +(7/2)x +3/4

    \[\begin{array}{*{35}{l}} <!-- /wp:paragraph --> <!-- wp:paragraph --> 2x2\text{ }+\left( 7/2 \right)x\text{ }+3/4 \\ <!-- /wp:paragraph --> <!-- wp:paragraph --> ~ \\ <!-- /wp:paragraph --> <!-- wp:paragraph --> \end{array}\]

The condition can likewise be composed as,

    \[8x2+14x+3\]

Parting the center term, we get,

    \[8x2+12x+2x+3\]

Taking the normal factors out, we get,

    \[4x\text{ }\left( 2x+3 \right)\text{ }+1\left( 2x+3 \right)\]

On gathering, we get,

    \[\left( 4x+1 \right)\left( 2x+3 \right)\]

In this way, the zeroes are,

    \[4x+1=0\Rightarrow x\text{ }=\text{ }-\text{ }1/4\]

    \[2x+3=0\Rightarrow x\text{ }=\text{ }-\text{ }3/2\]

Thusly, zeroes are – 1/4 and – 3/2

Check:

Amount of the zeroes = – (coefficient of x) ÷ coefficient of x2

    \[\alpha \text{ }+\text{ }\beta \text{ }=\text{ }\text{ }b/a\]

    \[\left( -\text{ }3/2 \right)\text{ }+\text{ }\left( -\text{ }1/4 \right)\text{ }=\text{ }\text{ }\left( 7 \right)/4\]

    \[=\text{ }\text{ }7/4\text{ }=\text{ }\text{ }7/4\]

Result of the zeroes = steady term ÷ coefficient of x2

    \[\alpha \text{ }\beta \text{ }=\text{ }c/a\]

    \[\left( -\text{ }3/2 \right)\left( -\text{ }1/4 \right)\text{ }=\text{ }\left( 3/4 \right)/2\]

    \[3/8\text{ }=\text{ }3/8\]

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