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This is one of our many quizzes on FOIL method. Click on this to access more such competitive quizzes on Algebra.

FOIL method Quiz 2

mathematics

 Solve the given question $(x - 3)3(2x - 1) = 0$ using FOIL method and get the value of $x$

 Use the  formula $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$ for finding the value of $x$.

 Applying FOIL method, $ \\ (x - 3)3(2x - 1) = 0 \\   \Rightarrow (x - 3)(6x - 3) = 0 \\  \Rightarrow 6{x^2} - 21x + 9 = 0 \\   \Rightarrow 2{x^2} - 7x + 3 = 0 \\ $ Using the formula $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$ for finding the value of $x$. we get, $ \\ x = \dfrac{{ - ( - 7) \pm \sqrt {{{( - 7)}^2} - 4(2)(3)} }}{{2(2)}} \\ \Rightarrow x = \dfrac{{7 \pm \sqrt {13} }}{4} \\  \Rightarrow x = 2.65,0.85 $

 Choose the correct answer for $(2x - 3)(\dfrac{2}{3}x - 2) = 0$ using FOIL method

 Multiply each term starting within brackets then multiply the terms outside the brackets.

 Simplifying the given equation using FOIL method: $ \\  (2x - 3)\left(\dfrac{2}{3}x - 2\right) = 0 \\   \Rightarrow \dfrac{{4{x^2}}}{3} - 4x - 2x + 6 = 0 \\   \Rightarrow \dfrac{{4{x^2}}}{3} - 6x + 6 = 0 \\   \Rightarrow \dfrac{{2{x^2}}}{3} - 3x + 3 = 0 \\   \Rightarrow 2{x^2} - 9x + 9 = 0 $

 Using the FOIL method, choose the correct answer for the value of $x$ using $7(x - 2)3(x - 3) = 0$

 Simplify the equation first then use the formula $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$ for finding the value of $x$.	 

 $  7(x - 2)3(x - 3) = 0 \\   \Rightarrow (7x - 6)(3x - 9) = 0 \\   \Rightarrow 21{x^2} - 63x - 18x + 54 = 0 \\   \Rightarrow 21{x^2} - 81x + 54 = 0 \\   \Rightarrow 7{x^2} - 27x + 18 = 0 \\  $ Now, to find $x$ we use, $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}} \\  x = \dfrac{{ - ( - 27) \pm \sqrt {{{( - 27)}^2} - 4(7)(18)} }}{{2(7)}} \\  x = \dfrac{{27 \pm 15}}{{14}} \\  x = 3,0.857 $

 Solve $(\sin x + \cos x)(\sin x + \cos x) = 0$ using FOIL method.

 Applying the trigonometric relation ${\sin ^2}x + {\cos ^2}x = 1$ and $2\sin x\cos x = \sin 2x$.

 $(\sin x + \cos x)(\sin x + \cos x) = 0 \\   \Rightarrow {\sin ^2}x + \sin x\cos x + \sin x\cos x + {\cos ^2}x = 0 \\   \Rightarrow {\sin ^2}x + 2\sin x\cos x + {\cos ^2}x = 0 \\ $ Applying the trigonometric relation ${\sin ^2}x + {\cos ^2}x = 1$ and $2\sin x\cos x = \sin 2x$ we get, $ \\ 1 + \sin 2x = 0 \\   \Rightarrow \sin 2x =  - 1 $

 Choose the correct option for $\dfrac{{(x - 3)(2x - 1)}}{{(x - 1)(x - 2)}} = 1$ using FOIL method.

 Simplify the equation 1st then use the  formula $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$ for finding the value of $x$.

 $\dfrac{{(x - 3)(2x - 1)}}{{(x - 1)(x - 2)}} = 1 \\   \Rightarrow (x - 3)(2x - 1) = (x - 1)(x - 2) \\  $ Applying the FOIL method: $ \\  (x - 3)(2x - 1) = (x - 1)(x - 2) \\   \Rightarrow 2{x^2} - x - 6x + 3 = {x^2} - 2x - x + 2 \\   \Rightarrow {x^2} - 4x + 1 = 0 \\  $ Now to find $x$ we use $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}} \\   x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}} \\   \Rightarrow x = \dfrac{{ - ( - 4) \pm \sqrt {{{( - 4)}^2} - 4(1)(1)} }}{{2(1)}} \\   \Rightarrow x = \dfrac{{4 \pm \sqrt {12} }}{2} \\   \Rightarrow x = 3.73,0.26 $