Yes, the work is accurate.
The table outlines the application of the Zero Product Property, which states that if the product of two factors is zero, then at least one of the factors must be zero.
The original equation is given as (x+3)(x+4) = 0. The Zero Product Property dictates that we set each factor equal to zero to solve for x.
The table shows the correct application of this property with the following steps:
1. (x+3)(x+4) = 0 - The original quadratic equation.
2. x+3 = 0 or x+4 = 0 - The factors are each set equal to zero.
3. x = -3 or x = -4 - The solutions are found by solving each equation for x.
Therefore, Oliver's solutions are x = -3 and x = -4, which are consistent with the correct application of the Zero Product Property. There's no need for a third column setting x = 0, as that is not how the Zero Product Property is used in this context. Oliver's work is indeed correct.
Use the table to answer the question.%0D%0A%0D%0A(x+3)(x+4)=0 %0D%0Ax−3=0 %09 x+4=0 %0D%0Ax=3 %09 x=−4 While using the Zero Product Property to find the values of x that make the quadratic equation (x+3)(x+4) equals 0, Oliver completed the work provided in the table. Is Oliver’s work accurate?(1 point)%0D%0AResponses%0D%0A%0D%0ANo, x+4=0 in the second row should be x−4=0.%0D%0ANo, x plus 4 equals 0 in the second row should be x minus 4 equals 0 .%0D%0A%0D%0ANo, x−3=0 in the second row should be x+3=0.%0D%0ANo, x minus 3 equals 0 in the second row should be x plus 3 equals 0 .%0D%0A%0D%0ANo, he should have a third column where he should set x=0.%0D%0ANo, he should have a third column where he should set x equals 0 .%0D%0A%0D%0AYes, the work is accurate.%0D%0AYes, the work is accurate.%0D%0ASkip to navigation%0D%0A%0D%0A%0D%0AItem%0D%0A1%0D%0A, Unattempted.%0D%0AItem%0D%0A2%0D%0A, Unattempted.%0D%0AItem%0D%0A3%0D%0A, Unattempted.%0D%0AItem%0D%0A4%0D%0A, Unattempted.%0D%0AItem%0D%0A5%0D%0A, Unattempted.%0D%0A%0D%0A%0D%0A%0D%0APreviousNext
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