Question

Describe and correct the error a student made in factoring 2x^2 + 1 1 x + 15.
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Answer 1

(2x + 5)(x+3) = 0

Explanation:

ac = 5 * 6 = 30
b = 5 + 6 = 11
Factorizing 2x² + 11x + 15,
2x² + 6x + 5x + 15 = 0
2x(x + 3) + 5(x+3) = 0
(2x + 5)(x+3) = 0

∴ x =
`(-5)/(2)`, -3

(There was a mistake probably because the quadratic equation was considered with 1 as the coefficient as x², so the factorized equation was for x² + 11x + 30. You must take the coefficient of x² as 2 instead, according to the given equation.)

Answer 2

The student forgot to divide the factors, 5 and 6, by the "a" value. In this case, that is 2. Then, simplify the fractions. Once we have these values, the numerator is the constant value and the denominator is the coefficient.

5/2 = 5/2 ➜ (2x + 5)

6/2 = 3 ➜ (x + 3)

This student should have factored the expression as;

(2x + 5)(x + 3)