Question

Verify if 1/2 and-3/2 are the polynomial 8 x cube minus 4 x square minus 18 x + 9 factorise polynomial

Answer 1

Given:-

$\bullet \;\;\;\;\normalsize\sf f(x) = 8x^3 - 4x^2 - 18x + 9$

To Verify:-

$\bullet \;\;\;\;\normalsize\sf \dfrac{1}{2}\;amd\; \dfrac{-3}{2}\;are\;the\;zeroes\;of\;the\;polynomial$

Solution:-

✫ Let $\normalsize\sf x = \dfrac{1}{2}$

$\;\;\;\;\;\;\;\;\small{\underline{\sf{\purple{\dag\;Putting\;value\;of\;x\;as\;\dfrac{1}{2} in:-}}}}$

$\normalsize\sf {\underline{F(x) = 8x^3 - 4x^2 - 18x + 9}}$

$\star\;\;\;\normalsize\sf f \bigg( \dfrac{1}{2} \bigg) = 8 \bigg( \dfrac{1}{2} \bigg)^3 - 4 \bigg( \dfrac{1}{2} \bigg)^2 - 18 \bigg( \dfrac{1}{2} \bigg) + 9 \\\\ \dashrightarrow\normalsize\sf \cancel{8} \bigg( \dfrac{1}{ \cancel{8}} \bigg) - \cancel{4} \bigg( \dfrac{1}{ \cancel{4}} \bigg) - \cancel{18} \bigg( \dfrac{1}{ \cancel{2}} \bigg) + 9 \\\\ \dashrightarrow\normalsize\sf \cancel{1} \cancel{- 1} \cancel{- 9 } \cancel{+ 9} \\\\ \dashrightarrow\normalsize\sf 0$

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✫ Let $\normalsize\sf x = \dfrac{-3}{2}$

$\;\;\;\;\;\;\;\;\small{\underline{\sf{\purple{ \dag\;Putting\;value\;of\;x\;as\;\dfrac{1}{2} in:-}}}}$

$\normalsize\sf {\underline{F(x) = 8x^3 - 4x^2 - 18x + 9}}$

$\star\;\;\;\normalsize\sf f \bigg( \dfrac{-3}{2} \bigg) = 8 \bigg( \dfrac{-3}{2} \bigg)^3 - 4 \bigg( \dfrac{-3}{2} \bigg)^2 - 18 \bigg( \dfrac{-3}{2} \bigg) + 9 \\\\ \dashrightarrow\normalsize\sf \cancel{8} \bigg( \dfrac{-27}{ \cancel{8}} \bigg) - \cancel{4} \bigg( \dfrac{9}{ \cancel{4}} \bigg) - \cancel{18} \bigg( \dfrac{-3}{ \cancel{2}} \bigg) + 9 \\\\ \dashrightarrow\normalsize\sf \cancel{-27} \cancel{-9 } \cancel{+ 27 } \cancel{+ 9} \\\\ \dashrightarrow\normalsize\sf 0$

$\therefore\sf \dfrac{1}{2}\;amd\; \dfrac{-3}{2}\;are\;the\;zeroes\;of\;the\;given\;polynomial\;f(x)$

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$\normalsize\sf\;\;\;\;\;\; \dag\; TO\;FACTORIZE:-$

$\normalsize\sf \star\;{\underline{f(x) = 8x^3 - 4x^2 - 18x + 9}} \\\\ \dashrightarrow\normalsize\sf 4x^2(2x - 1) -9(2x - 1) \\\\ \dashrightarrow\normalsize{\underline{\sf{\red{(4x^2 - 9)(2x - 1)}}}}$

$\;\;\;\;\;\;\;\;\;\normalsize {\underline{\underline{\sf{\dag\;Hence\;Solved!}}}}$

$\rule{200}{3}$