Question

Find the values of p,q, and r

(px + q) (2x + 5) = 6x² + 11x + r

please show your work .

Answer 1

so hmmmm recall FOIL.

on a two binomial set

(ax+b)(cx+d), you get a trinomial that is

(a*c)x² + [(ax*d) + (cx*b)] x +(b*d).

that said, then


`\bf (px+q)(2x+5)=\stackrel{px\cdot 2x}{6x^2}+\stackrel{(px\cdot 5)+(2x\cdot q)}{11x}+r \\\\\\ therefore \\\\\\ px\cdot 2x=6x^2\implies 2px^2=6x^2\implies p=\cfrac{6x^2}{2x^2}\implies \boxed{p=3}\\\\ -------------------------------\\\\ (px\cdot 5)+(2x\cdot q)=11x\implies 5px+2qx=11x \\\\\\ 5\left( \boxed{3} \right)x+2qx=11x\implies 15x+2qx=11x\implies 2qx=-4x \\\\\\ q=\cfrac{-4x}{2x}\implies \boxed{q=-2}`

and "r" of course, is just (q*5), and surely you know what that is.

Answer 2

(px+q)(2x+5)=6x^2+11x+r

2px² + (5p + 2q)x + 5q = 6x² + 11x + r

Start with the x² coefficients:
2px² = 6x²
2p = 6
p = 3

Next, the x coefficients:
(5p + 2q)x = 11x
( (5)(3) + 2q) = 11
15 + 2q = 11
2q = -4
q = -2

Next, the constant terms:
5q = r
5(-2) = r
r = -10

Thus,
p = 3, q = -2, r = -10