if the reminder of deviation of x cube + 2 x square + kx + 3 by x minus 3 is 21 find the person then the value of tea
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hii mate!!
We know that,
Dividend=Divisor×Quotient+Remainder
Given that,
Dividend=x³+2x²+kx+3
Divisor=x-3
Remainder=21
Now we have,
x³+2x²+kx+3=(x-3)quotient+21...............1
(x³+2x²+kx-18)/(x-3)=quotient
Now the remainder will be equal to 0.
From above picture we get,
Remainder=0
3(k+15)-18=0
k+15=6
k=-9
Now eq1 we have,
x³+2x²-9x+3=(x-3)(x²+5x+6)+21
x³+2x²-9x-18=(x-3)(x²+2x+3x+6)
x³+2x²-9x-18=(x-3){x(x+2)+3(x+2)}
x³+2x²-9x-18=(x-3)(x+3)(x+2)
Therefore zeroes are 3 , -3 and -2
Hence value of k=-9
and zeroes of given cubic polynomial are 3 , -3 and -2.
hope it helps !!☺☺☺☺☺☺☺
We know that,
Dividend=Divisor×Quotient+Remainder
Given that,
Dividend=x³+2x²+kx+3
Divisor=x-3
Remainder=21
Now we have,
x³+2x²+kx+3=(x-3)quotient+21...............1
(x³+2x²+kx-18)/(x-3)=quotient
Now the remainder will be equal to 0.
From above picture we get,
Remainder=0
3(k+15)-18=0
k+15=6
k=-9
Now eq1 we have,
x³+2x²-9x+3=(x-3)(x²+5x+6)+21
x³+2x²-9x-18=(x-3)(x²+2x+3x+6)
x³+2x²-9x-18=(x-3){x(x+2)+3(x+2)}
x³+2x²-9x-18=(x-3)(x+3)(x+2)
Therefore zeroes are 3 , -3 and -2
Hence value of k=-9
and zeroes of given cubic polynomial are 3 , -3 and -2.
hope it helps !!☺☺☺☺☺☺☺
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