Respuesta :
The two possible values for c are 31 and 41
Step-by-step explanation:
Given Expression:
[tex](a x+2)(b x+7)=15 x^{2}+c x+14[/tex]
a + b = 8
To expand,
Multiply a x with (b x + 7) = [tex]a b x^{2}+7 a x[/tex]
Multiply 2 with (b x + 7) = 2 b x +14
Now, combining the above, we get
[tex]a b x^{2}+7 a x+2 b x+14=15 x^{2}+c x+14[/tex]
[tex]a b x^{2}+x(7 a+2 b)+14=a b x^{2}[/tex]
When comparing both sides, we get
a b = 15, 7 a + 2 b = c
[tex]a=\frac{15}{b} \text { or } b=\frac{a}{15}[/tex]
Now, substitute above value in a + b = 8. So,
[tex]\frac{15}{b}+b=8[/tex]
[tex]15+b^{2}=8 b[/tex]
[tex]b^{2}-8 b+15=0[/tex]
Factorising above, we get the equation as
[tex]b^{2}-3 b-5 b+15=0[/tex]
(b - 3) (b - 5) = 0
b = 3 and 5
If b = 3, then
[tex]a=\frac{15}{b}=\frac{15}{3}=5[/tex]
If b = 5, then
[tex]a=\frac{15}{5}=3[/tex]
If a = 3, b = 5
c =7 a + 2 b = 7 (3) + 2 (5) = 21 + 10 = 31
If a = 5, b = 3
c =7 (5) +2 (3) = 35 + 6 = 41
Therefore, the values of ‘c’ are 31 and 41 .
