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Find the zeroes of the following polynomials by factorisation method. t3 – 2t2 – 15t
CBSE | Class 10 | Exercise 2.3 | Maths | NCERT Exemplar | Polynomials
Find the zeroes of the following polynomials by factorisation method. t3 – 2t2 – 15t

    \[t3\text{ }\text{ }2t2\text{ }\text{ }15t\]

Taking t normal, we get,

    \[t\text{ }\left( \text{ }t2\text{ }-\text{ }2t\text{ }-\text{ }15 \right)\]

Parting the center term of the situation t2 – 2t – 15, we get,

    \[t\left( \text{ }t2\text{ }-\text{ }5t\text{ }+\text{ }3t\text{ }-\text{ }15 \right)\]

Taking the normal factors out, we get,

    \[(t\text{ }\left( t-5 \right)\text{ }+3\left( t-5 \right)\]

On gathering, we get,

    \[t\text{ }\left( t+3 \right)\left( t-5 \right)\]

In this way, the zeroes are,

t=0

t+3=0 ⇒ t= – 3

t – 5=0 ⇒ t=5

Subsequently, zeroes are 0, 5 and – 3

Check:

Amount of the zeroes = – (coefficient of x2) ÷ coefficient of x3

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

    \[\left( 0 \right)\text{ }+\text{ }\left( -\text{ }3 \right)\text{ }+\text{ }\left( 5 \right)\text{ }=\text{ }\text{ }\left( -\text{ }2 \right)/1\]

= 2 = 2

Amount of the results of two zeroes at an at once of x ÷ coefficient of x3

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

    \[\left( 0 \right)\left( -\text{ }3 \right)\text{ }+\text{ }\left( -\text{ }3 \right)\text{ }\left( 5 \right)\text{ }+\text{ }\left( 0 \right)\text{ }\left( 5 \right)\text{ }=\text{ }\text{ }15/1\]

= – 15 = – 15

Result of all the zeroes = – (consistent term) ÷ coefficient of x3

    \[\alpha \beta \gamma \text{ }=\text{ }\text{ }d/a\]

    \[\left( 0 \right)\left( -\text{ }3 \right)\left( 5 \right)\text{ }=\text{ }0\]

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