# Equations- Factorization of a polynomial equation – Exercise 5602

Exercise

$$4p^2-21p-18=0$$

$$(p-6)(4p+3)$$

Solution

$$4p^2-21p-18=0$$

It’s a quadratic equation. Our equation coefficients are

$$a=4, b=-21, c=-18$$

We solve it with the quadratic formula. Putting the coefficients in the formula gives us

$$p_{1,2}=\frac{21\pm \sqrt{{21}^2-4\cdot 4\cdot (-18)}}{2\cdot 4}=$$

$$=\frac{21\pm \sqrt{729}}{8}=$$

$$=\frac{21\pm 27}{8}$$

Hence, we get the solutions:

$$p_1=\frac{21+27}{8}=6$$

$$p_2=\frac{21-27}{8}=-\frac{6}{8}=-\frac{3}{4}$$

Thus, the factorizing of the quadratic equation is

$$4p^2-21p-18=$$

$$=4(p-6)(p+\frac{3}{4})=$$

$$=(p-6)(4p+3)$$

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