Factorization of the given polynomial x3−7x+6 will be ____.

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Factorization of the given polynomial x3−7x+6 will be ____.

Infinity Learn · Accepted Answer

Given polynomial, x3−7x+6Let us take the given polynomial as p(x)=x3−7x+6Let us consider all the factors of the 6.By using trial method-Let us substitute, −1 in the place of x. We getp(−1) = (−1)3−7(−1)+6= −1−7+6= −2 ≠ 0 Therefore, (x+1) is not the factor of a given polynomial.Again let us substitute 1 in the place of x . We get,p(1) = (1)3−7(1)+6 = 1−7+6 = 0 Therefore, (x−1) is a factor of the given polynomial.Now, let us find the quotient on dividing x3−7x+6 by (x−1)This can be solved by using method of long divisionWe know that ,Dividend = divisor × quotient + remainderTherefore , x3−7x+6=(x−1)×(x2+x−6)+(0)x3−7x+6=(x−1)×(x2+x−6).....(1) Let us consider q(x)=x2+x−6Let us find the factors of the q(x)=x2+x−6 by using factorization methodComparing q(x)=x2+x−6with expression with ax2+bx+c.We get a=1,b=1,c=−6Now multiply “a” term with “c” term and factor the product of (ac) such that the sum of factors equals to b.The product of a, c is 1(−6) = −6. Factor -6 as -2,3 such that the sum of factors (-2)+3 = 1 which equals a coefficient of x that is b.That is x2+x−6 = x2+3x−2x−6In the next step , we are taking x as common in first two terms and -2 as common in next two terms= x(x+3) − 2(x+3)Now taking (x+3) from above equation= (x+3)(x−2)Now substitute above factors in equation (1) , we get x3−7x+6 = (x−1)×[(x+3)×(x−2)]= (x−1)(x+3)(x−2) Thus, the factors of the given polynomial are (x−1)(x+3)(x−2).

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