231
Question
If two zeroes of the polynomial f(x)=x3−4x2−3x+12 are √3,−√3 then find its third zero.
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Solution
Since, it is given that √3 and −√3 are the zeroes of the polynomial f(x)=x3−4x2−3x+12, therefore, (x−√3) and (x+√3) are also the zeroes of the given polynomial. Now, consider the product of zeroes as follows:
(x−√3)(x+√3)=(x)2−(√3)2(∵a2−b2=(a+b)(a−b))=x2−3
We now divide x3−4x2−3x+12 by (x2−3) as shown in the above image:
From the division, we observe that the quotient is x−4 and the remainder is 0.
Since x−4 is the quotient,
Hence, the third zero of f(x)=x3−4x2−3x+12 is 4.
Since, it is given that √3 and −√3 are the zeroes of the polynomial f(x)=x3−4x2−3x+12, therefore, (x−√3) and (x+√3) are also the zeroes of the given polynomial. Now, consider the product of zeroes as follows:
(x−√3)(x+√3)=(x)2−(√3)2(∵a2−b2=(a+b)(a−b))=x2−3
We now divide x3−4x2−3x+12 by (x2−3) as shown in the above image:
From the division, we observe that the quotient is x−4 and the remainder is 0.
Since x−4 is the quotient,
Hence, the third zero of f(x)=x3−4x2−3x+12 is 4.
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