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Mathematics

Using the remainder and factor theorems, factorize the polynomial.

x3 + 10x2 - 37x + 26

Factorisation

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Answer

Let, f(x) = x3 + 10x2 - 37x + 26.

Substituting, x = 1 in f(x) we get :

f(1) = (1)3 + 10(1)2 - 37(1) + 26

= 1 + 10 - 37(1) + 26

= 37 - 37

= 0.

Since, f(1) = 0, thus (x − 1) is a factor of f(x).

Dividing, x3 + 10x2 - 37x + 26 by (x - 1), we get :

x]3)x2+11x26x1)x3+10x237x+26x2x3+x2x2x,,,311x237xxl2fx3] +11x2+11xx2]euo[ki]x3okk 26x+26x2x3o;llk]lmttk +26x+26x2x,jok2xmm2 9x×\begin{array}{l} \phantom{x - ]3)}{x^2 + 11x - 26} \ x - 1\overline{\smash{\big)}x^3 + 10x^2 - 37x + 26} \ \phantom{x - 2}\underline{\underset{-}{}x^3 \underset{+}{-}x^2} \ \phantom{{x - 2}x^,,,3-}11x^2 - 37x \ \phantom{{x -l2}fx^3]\space}\underline{\underset{-}{+}11x^2 \underset{+}{-}11x} \ \phantom{{x - 2]euo[ki]}x^3okk\space}{-26x + 26} \ \phantom{{x - 2}x^3o;llk]lmttk\space}\underline{\underset{+}{-}26x\underset{-}{+}26} \ \phantom{{x - 2}{x^,jo-k2x^mm2\space}{-9x}}\times \end{array}

∴ x3 + 10x2 - 37x + 26 = (x - 1)(x2 + 11x - 26)

= (x - 1)(x2 + 13x - 2x - 26)

= (x - 1)[x(x + 13) - 2(x + 13)]

= (x - 1)(x + 13)(x - 2).

Hence, x3 + 10x2 - 37x + 26 = (x - 1)(x + 13)(x - 2).

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