Question

find the zeros of plolynomialp(x)=4x2+7x+3 by factoring it and verify the relationship between the zeros and coffients of p(x)​

Answer 1

Step-by-step explanation:

$4 {x}^{2} + 7x + 3 \\ 4 {x}^{2} + 4x + 3x + 3 = 0 \\ 4x(x + 1) + 3(x + 1) = 0 \\ (x + 1)(4x + 3) = 0 \\ x + 1 = 0 \: \: \: \: \: \: 4x + 3 = 0 \\ x = - 1 \: \: \: \: \: \: \: \: 4x = - 3 \\ x = - 1 \: \: \: \: \: \: \: \: \: x = - \frac{3}{4}$

sum of zeroes = -b/a = -7/4

$= - 1 + ( - \frac{3}{4} ) \\ = - 1 - \frac{3}{4} \\ = \frac{ - 4 - 3}{4} \\ = - \frac{7}{4}$

product of zeroes = c/a = 3/4

$= - 1 \times - \frac{3}{4} \\ = \frac{3}{4}$

Hope you get your answer