To solve the trinomial \(4x^2 - 17x + 15\), we can factor it. We need to find two numbers that multiply to \(4 \cdot 15 = 60\) (the product of the leading coefficient and the constant) and add to \(-17\) (the coefficient of \(x\)).
Let's find the pair of numbers:
The pairs of factors of 60 are:
- (1, 60)
- (2, 30)
- (3, 20)
- (4, 15)
- (5, 12)
- (6, 10)
We look for a pair that adds to \(-17\):
- The pair \(-12\) and \(-5\) works since \(-12 + (-5) = -17\).
Now we can rewrite the middle term of the trinomial using \(-12\) and \(-5\):
\[ 4x^2 - 12x - 5x + 15 \]
Next, we can group the terms:
\[ (4x^2 - 12x) + (-5x + 15) \]
Factoring each group gives us:
\[ 4x(x - 3) - 5(x - 3) \]
Now we can factor out \((x - 3)\):
\[ (4x - 5)(x - 3) \]
So the factored form of the trinomial is:
\[ (4x - 5)(x - 3) = 0 \]
Now, we can set each factor equal to zero to find the solutions:
-
\(4x - 5 = 0\) \[ 4x = 5 \ x = \frac{5}{4} \]
-
\(x - 3 = 0\) \[ x = 3 \]
Thus, the solutions to the equation \(4x^2 - 17x + 15 = 0\) are:
\[ x = \frac{5}{4} \quad \text{and} \quad x = 3 \]
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