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( 5-5)( 5+2)? =-6 & ⇔ 0 ≠-6 * ( -2-5)( -2+2)? =-6 & ⇔ 0 ≠-6 * As we can see, the proposed solutions does not result in true statements. Therefore, these values of x cannot be solutions to the equation.
Multiply parentheses
Subtract term
LHS+6=RHS+6
To fill in the remaining two corners, we need two x-terms that have a sum of -3x and a product of - 4x^2.
Notice that the product is negative. This means one factor must be positive and the other must be negative. |c|c|c|r|c| [-1em] Product & ax(bx) & ax+bx & Sum & -3x? [0.2em] [-1em] -4x^2 & - x(4x) &- x+4x& 3x & * [0.1em] -4x^2 & - 4x(x) & -4x+x& -3x & âś“ [0.1em] When one term is - 4x and the other is x, we have a product of -4x^2 and a sum of -3x. Now we can complete the diamond and generic rectangle.
To factor the left-hand side we add each side of the generic rectangle and multiply the sums. x^2-3x-4=0 ⇓ (x+1)(x-4)=0
Use the Zero Product Property
(I): LHS-1=RHS-1
(II): LHS+4=RHS+4