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((0 - 5x2) - 35x) - 50 3.1 Pull out like factors : -5x2 - 35x - 50 = -5 • (x2 + 7x + 10) Trying to factor by splitting the middle term3.2 Factoring x2 + 7x + 10 The first term is, x2 its coefficient is 1 . The middle term is, +7x its coefficient is 7 . The last term, "the constant", is +10 Step-1 : Multiply the coefficient of the first term by the constant 1 • 10 = 10 Step-2 : Find two factors of 10 whose sum equals the coefficient of the middle term, which is 7 .
x • (x+2) Add up the last 2 terms, pulling out common factors :5 • (x+2) Step-5 : Add up the four terms of step 4 :(x+5) • (x+2) Which is the desired factorization Final result :-5 • (x + 5) • (x + 2)
Changes made to your input should not affect the solution: (1): "x5" was replaced by "x^5". Step by step solution :Step 1 :Equation at the end of step 1 :(5 • (x2)) - (5•7x50) = 0Equation at the end of step 2 :5x2 - (5•7x50) = 0Step 3 :Step 4 :Pulling out like terms :4.1 Pull out like factors : 5x2 - 35x50 = -5x2 • (7x48 - 1) Trying to factor as a Difference of Squares :4.2 Factoring: 7x48 - 1 Theory : A difference of two perfect squares, A2 - B2 can be factored into Proof : (A+B) • (A-B) = Note : - AB + AB equals zero and is therefore eliminated from the expression. Check :Ruling : Binomial can not be factored as the difference of two perfect squaresTrying to factor as a Difference of Cubes:4.3 Factoring: 7x48 - 1 Theory : A difference of two perfect cubes, a3 - b3 can be factored into (a-b) • (a2 +ab +b2) Proof : (a-b)•(a2+ab+b2) = Check : 7 is not a cube !! Ruling : Binomial can not be factored as the difference of two perfect cubes Equation at the end of step 4 :-5x2 • (7x48 - 1) = 0Step 5 :Theory - Roots of a product :5.1 A product of several terms equals zero.When a product of two or more terms equals zero, then at least one of the terms must be zero.We shall now solve each term = 0 separatelyIn other words, we are going to solve as many equations as there are terms in the productAny solution of term = 0 solves product = 0 as well. Solving a Single Variable Equation :5.2 Solve : -5x2 = 0Multiply both sides of the equation by (-1) : 5x2 = 0 Divide both sides of the equation by 5: x2 = 0 When two things are equal, their square roots are equal. Taking the square root of the two sides of the equation we get: x = ± √ 0 Any root of zero is zero. This equation has one solution which is x = 0 Solving a Single Variable Equation : 5.3 Solve : 7x48-1 = 0Add 1 to both sides of the equation : x48 = 1/7 = 0.143 x = 48th root of (1/7)The equation has two real solutions These solutions are x = 48th root of ( 0.143) = ± 0.96027 Three solutions were found :
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