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Question

If the zeros of the polynomial (x46x326x2+138x35) are (2+3) and (23), find the other zeros.

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Solution


Since, it is given that 2+3 and 23 are the zeroes of the polynomial p(x)=x46x326x2+138x35, therefore, (x23) and (x2+3) are also the zeroes of the given polynomial. Now, consider the product of zeroes as follows: Now, consider the product of zeroes as follows:

(x23)(x2+3)[(x2)3][(x2)+3]=(x2)2(3)2(a2b2=(a+b)(ab))=[x2+22(2×x×2)]3((ab)2=a2+b22ab)=x2+44x3=x24x+1

We now divide x46x326x2+138x35 by (x24x+1) as shown in the below image:



From the division, we observe that the quotient is x22x35 and the remainder is 0.

Now, we factorize the quotient x22x35 by equating it to 0 to find the other zeroes of the given polynomial:

x22x35=0x27x+5x35=0x(x7)+5(x7)=0(x+5)(x7)=0x=5,x=7

Hence, the other two zeroes of x46x326x2+138x35 are 5,7.

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