To find the zeros of the function \( y = 2x^2 + 5x - 7 \), we can either graph the function or use the quadratic formula. Let's first find the zeros using the quadratic formula:
The quadratic formula is given by:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
For the function \( y = 2x^2 + 5x - 7 \):
- \( a = 2 \)
- \( b = 5 \)
- \( c = -7 \)
- Calculate the discriminant \( b^2 - 4ac \):
\[ b^2 - 4ac = 5^2 - 4 \cdot 2 \cdot (-7) = 25 + 56 = 81 \]
- Substitute the values into the quadratic formula:
\[ x = \frac{-5 \pm \sqrt{81}}{2 \cdot 2} \]
\[ x = \frac{-5 \pm 9}{4} \]
This gives us two solutions:
- For \( x = \frac{-5 + 9}{4} = \frac{4}{4} = 1 \)
- For \( x = \frac{-5 - 9}{4} = \frac{-14}{4} = -\frac{7}{2} \)
Thus, the zeros of the function are:
\[ 1, -\frac{7}{2} \]
Now, to rewrite the function in factored form, we can use the zeros we found. The factored form of a quadratic function is given as:
\[ y = a(x - r_1)(x - r_2) \]
Where \( r_1 \) and \( r_2 \) are the roots (zeros) of the function. Here \( r_1 = 1 \) and \( r_2 = -\frac{7}{2} \). Therefore, we express \( y \) as:
\[ y = 2(x - 1)\left(x + \frac{7}{2}\right) \]
To clear the fraction in \( x + \frac{7}{2} \), multiply by 2:
\[ y = 2(x - 1)\left(\frac{2x + 7}{2}\right) = (x - 1)(2x + 7) \]
Thus, the final factored form is:
\[ y = 2(x - 1)\left(x + \frac{7}{2}\right) \] or simply
\[ y = 2\left(x - 1\right)\left(2x + 7\right) \]
So the final answers are:
- The zeros of the function: \( \boxed{1, -\frac{7}{2}} \)
- The factored form: \( y = 2(x - 1)\left(2x + 7\right) \)
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