Question

The product of $7d^2-3d+g$ and $3d^2+hd-8$ is $21d^4-44d^3-35d^2+14d-16$. What is $g+h$?

Answer 1

Answer:

g = 2  and h = -5

Step-by-step explanation:

To find the product with unknown values of variables, multiply the terms and equate the value to the given product.

$(7d^{2} - 3d + g)(3d^{2} + hd - 8)$ = $21d^{4} + 7hd^{3} - 56d^{2} - 9d^{3} - 3hd^{2} + 24d + 3gd^{2} + ghd - 8g$

$(7d^{2} - 3d + g)(3d^{2} + hd - 8)$     = $21d^{4} + (7h-9) d^{3} + (3g-56-3h)d^{2} + (24+gh)d -8g$   ... (1)

Comparing eq(1) to the product given in question,

-8g = -1

g = 2

24 + gh = 14, sub g =2,

24 + 2h = 14

h = -5