Question 84Given, 9𝑥2 - 4 is a factor of 9𝑥3 - m𝑥2 - n𝑥 + 8 :(a) find the value of m and n using the remainder and factor theorem.(b) factorise the given polynomial completely.

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(a) Simplifying 9𝑥2 - 4, we get :⇒ 9𝑥2 - 4⇒ (3𝑥)2 - 22⇒ (3𝑥 + 2)(3𝑥 - 2).We know that,If (x - a) is a factor of f(x), then f(a) = 0.Given,9𝑥2 - 4 is a factor of 9𝑥3 - m𝑥2 - n𝑥 + 8.∴ (3𝑥 + 2) and (3𝑥 - 2) are the factors of 9𝑥3 - m𝑥2 - n𝑥 + 8.⇒ 3𝑥 + 2 = 0⇒ 3𝑥 = -2⇒ 𝑥 = −23-\dfrac{2}{3}−32​Substituting x = −23-\dfrac{2}{3}−32​ in 9𝑥3 - mx2 - nx + 8, we get remainder = 0.⇒9×(−23)3−m×(−23)2−n×(−23)+8=0⇒9×−827−m×49+2n3+8=0⇒−83−4m9+2n3+8=0⇒−24−4m+6n+729=0⇒6n−4m+489=0⇒6n−4m+48=0⇒2(3n−2m+24)=0⇒3n−2m+24=0⇒3n−2m=−24 ..........(1)\Rightarrow 9 \times \Big(-\dfrac{2}{3}\Big)^3 - m \times \Big(-\dfrac{2}{3}\Big)^2 - n \times \Big(-\dfrac{2}{3}\Big) + 8 = 0 \[1em] \Rightarrow 9 \times -\dfrac{8}{27} - m \times \dfrac{4}{9} + \dfrac{2n}{3} + 8 = 0 \[1em] \Rightarrow -\dfrac{8}{3} - \dfrac{4m}{9} + \dfrac{2n}{3} + 8 = 0 \[1em] \Rightarrow \dfrac{-24 - 4m + 6n + 72}{9} = 0 \[1em] \Rightarrow \dfrac{6n - 4m + 48}{9} = 0 \[1em] \Rightarrow 6n - 4m + 48 = 0 \[1em] \Rightarrow 2(3n - 2m + 24) = 0 \[1em] \Rightarrow 3n - 2m + 24 = 0 \[1em] \Rightarrow 3n - 2m = -24 \text{ ..........(1)}⇒9×(−32​)3−m×(−32​)2−n×(−32​)+8=0⇒9×−278​−m×94​+32n​+8=0⇒−38​−94m​+32n​+8=0⇒9−24−4m+6n+72​=0⇒96n−4m+48​=0⇒6n−4m+48=0⇒2(3n−2m+24)=0⇒3n−2m+24=0⇒3n−2m=−24 ..........(1)⇒ 3𝑥 - 2 = 0⇒ 3𝑥 = 2⇒ 𝑥 = 23\dfrac{2}{3}32​Substituting 𝑥 = 23\dfrac{2}{3}32​ in 9𝑥3 - m𝑥2 - n𝑥 + 8, we get remainder = 0.⇒9×(23)3−m×(23)2−n×(23)+8=0⇒9×827−m×49−2n3+8=0⇒83−4m9−2n3+8=0⇒24−4m−6n+729=0⇒96−6n−4m9=0⇒96−6n−4m=0⇒4m+6n=96⇒2(2m+3n)=96⇒2m+3n=962⇒2m+3n=48 ..........(2)\Rightarrow 9 \times \Big(\dfrac{2}{3}\Big)^3 - m \times \Big(\dfrac{2}{3}\Big)^2 - n \times \Big(\dfrac{2}{3}\Big) + 8 = 0 \[1em] \Rightarrow 9 \times \dfrac{8}{27} - m \times \dfrac{4}{9} - \dfrac{2n}{3} + 8 = 0 \[1em] \Rightarrow \dfrac{8}{3} - \dfrac{4m}{9} - \dfrac{2n}{3} + 8 = 0 \[1em] \Rightarrow \dfrac{24 - 4m - 6n + 72}{9} = 0 \[1em] \Rightarrow \dfrac{96 - 6n - 4m}{9} = 0 \[1em] \Rightarrow 96 - 6n - 4m = 0 \[1em] \Rightarrow 4m + 6n = 96 \[1em] \Rightarrow 2(2m + 3n) = 96 \[1em] \Rightarrow 2m + 3n = \dfrac{96}{2} \[1em] \Rightarrow 2m + 3n = 48 \text{ ..........(2)}⇒9×(32​)3−m×(32​)2−n×(32​)+8=0⇒9×278​−m×94​−32n​+8=0⇒38​−94m​−32n​+8=0⇒924−4m−6n+72​=0⇒996−6n−4m​=0⇒96−6n−4m=0⇒4m+6n=96⇒2(2m+3n)=96⇒2m+3n=296​⇒2m+3n=48 ..........(2)Adding equation (1) and (2), we get :⇒ 3n - 2m + 2m + 3n = -24 + 48⇒ 6n = 24⇒ n = 246\dfrac{24}{6}624​⇒ n = 4.Substituting value of n in equation (1), we get :⇒ 3(4) - 2m = -24⇒ 12 - 2m = -24⇒ -2m = -24 - 12⇒ -2m = -36⇒ m = −36−2\dfrac{-36}{-2}−2−36​ = 18.Hence, m = 18 and n = 4.(b) Substituting value of m and n in 9𝑥3 - m𝑥2 - n𝑥 + 8, we get :9𝑥3 - 18𝑥2 - 4𝑥 + 8Dividing 9𝑥3 - 18𝑥2 - 4𝑥 + 8 by 9𝑥2 - 4, we get :9x2−4)x−29x2−4)9x3−18x2−4x+8‾9x2−4))+−9x3−18x2−+4x‾9x2−4+9x3−−18x2−4x+89x2−4)+9x3−−+18x2−4x+−8‾9x2−4)x3−2x2(31)x×\begin{array}{l} \phantom{9x^2 - 4)}{\quad x - 2} \ 9x^2 - 4\overline{\smash{\big)}\quad 9x^3 - 18x^2 - 4x + 8} \ \phantom{9x^2 - 4)}\phantom{)}\underline{\underset{-}{+}9x^3 \phantom{- 18x^2} \underset{+}{-}4x} \ \phantom{{9x^2 - 4}{+9x^3 - }}-18x^2 \phantom{- 4x} + 8 \ \phantom{{9x^2 - 4)}{+9x^3 - }}\underline{\underset{+}{-}18x^2 \phantom{- 4x} \underset{-}{+} 8} \ \phantom{{9x^2 - 4)}{x^3-2x^{2}(31)}{x}}\times \end{array}9x2−4)x−29x2−4)9x3−18x2−4x+8​9x2−4))−+​9x3−18x2+−​4x​9x2−4+9x3−−18x2−4x+89x2−4)+9x3−+−​18x2−4x−+​8​9x2−4)x3−2x2(31)x×​∴ 9𝑥3 - 18𝑥2 - 4𝑥 + 8 = (9𝑥2 - 4)(𝑥 - 2)= (3𝑥 + 2)(3𝑥 - 2)(𝑥 - 2).Hence, factors are (3𝑥 + 2), (3𝑥 - 2) and (𝑥 - 2).

Short Answer Questions 2 — Question 7

Question 84