Problem

Factor by grouping (sometimes called the ac-method). \[ 5 x^{2}+6 x-8 \] First, choose a form with appropriate signs. Then, fill in the blanks with numbers to be used for grouping. Finally, show the factorization. Form: $05 x^{2}+\square x+\square x-8$ $05 x^{2}+\square x-\square x-8$ $05 x^{2}-\square x+\square x-8$ $05 x^{2}-\square x-\square x-8$ Factorization: $\square$ Check $02023 \mathrm{Mc}$

Solution

Step 1 :First, we need to find two numbers that multiply to -40 (which is the product of 5 and -8) and add to 6. The two numbers that satisfy these conditions are 10 and -4.

Step 2 :We can rewrite the equation as \(5x^2 + 10x - 4x - 8\).

Step 3 :Now we can factor by grouping. The first group is \(5x^2 + 10x\) and the second group is \(-4x - 8\).

Step 4 :We can factor out a \(5x\) from the first group and a \(-4\) from the second group. This gives us \(5x(x + 2) - 4(x + 2)\).

Step 5 :Now we can factor out the common factor of \((x + 2)\) to get the final factorization.

Step 6 :The factorization of the given quadratic equation \(5x^2 + 6x - 8\) is \(\boxed{(x + 2)(5x - 4)}\).

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Source: https://solvelyapp.com/problems/fQMQjchvQy/

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