If the remainder on division of x3 + 2x2 +kx+3 by (x-3) is 21. Find the quotient and value of K. Find the zeros of cubic polynomial.
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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.
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.
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Similar questions
Given that p(3) = 21
ie, 27 + 18 + 3k + 3 = 21
3k = -27
k = -27 / 3 = -9
p(x) = x3 + 2x2 + kx - 18
= x3 + 2x2 - 9x - 18
when x = -2, p(-2) = 0
ie, x + 2 is a factor, dividing p(x) by x + 2 ,we get,
x2 - x - 6 = (x - 3)(x + 2)
Hence the zeros of x3 + 2x2 - 9x - 18 are -2 , -2 ,3