[tex](ax+2)(bx+7)=abx^2+(7a+2b)x+14=15x^2+cx+14[/tex]
For the two quadratics to match, the coefficients of corresponding terms must be equal, so that
[tex]\begin{cases}ab=15\\7a+2b=c\end{cases}[/tex]
From the given options, we are only allowing [tex]c[/tex] to be an integer, which means [tex]a,b[/tex] must also be integers. Then we have eight possible choices for [tex]a,b[/tex]:
1, 15
3, 5
5, 3
15, 1
plus the same combinations with negative signs on both numbers. But since the choices for [tex]c[/tex] are also positive, we can ignore the negative versions.
If [tex](a,b)=(1,15)[/tex], then [tex]7a+2b=37[/tex].
If [tex](a,b)=(3,5)[/tex], then [tex]7a+2b=31[/tex].
If [tex](a,b)=(5,3)[/tex], then [tex]7a+2b=41[/tex].
If [tex](a,b)=(15,1)[/tex], then [tex]7a+2b=107[/tex].
So the answer is C.
The two possible values of [tex]C[/tex] is option [tex]C[/tex] [tex](31[/tex] and [tex]41)[/tex].
Given,
[tex](ax+2)(bx+7)=15x^2+cx+14\\a+b=8[/tex]
Simplify the equation of LHS as,
[tex](ax+2)(bx+7)=abx^2+2bx+7ax+14\\=abx^2+x(7a+2b)+14[/tex]
Compare LHS and RHS as,
[tex]abx^2+x(7a+2b)+14=15x^2+cx+14\\abx^2+(7a+2b)x=15x^2+cx[/tex]
The two equations are:
[tex]ab=15\\7a+2b=c[/tex]
From the given,
[tex]a+b=8[/tex]
Subtract [tex]b[/tex] on both sides in the equation as,
[tex]a+b-b=8-b\\a=8-b[/tex]
Substitute [tex]a=8-b[/tex] in the equation [tex]ab=15[/tex] as,
[tex](8-b)b=15\\8b-b^2=15\\-b^2+8b=15\\[/tex]
Solve the equation by setting equal to zero as,
[tex]-b^2+8b-15=0\\b^2-8b+15=0[/tex]
Factorize the equation as follows:
[tex]b^2--8b+15=0\\b^2-5b-3b+15=0\\b(b-5)-3(b-5)=0\\(b-3)(b-5)=0[/tex]
Solve the factors as,
[tex]b-3=0\\b=3[/tex]
And
[tex]b-5=0\\b=5[/tex]
Substitute [tex]b=3[/tex] in [tex]a=8-b[/tex] as,
[tex]a=8-3\\=5[/tex]
Thus, [tex]a=5,b=3[/tex].
Substitute [tex]b=5[/tex] in [tex]a=8-b[/tex] as,
[tex]a=8-5\\=3[/tex]
Thus, [tex]a=3,b=5[/tex].
Substitute [tex]a=3,b=5[/tex] in the equation [tex]7a+2b=c[/tex] as,
[tex]7(3)+2(5)=c\\21+10=c\\c=31[/tex]
Substitute [tex]a=5,b=3[/tex] in the equation [tex]7a+2b=c[/tex] as,
[tex]7(5)+2(3)=c\\35+6=c\\c=41[/tex]
Thus, the two possible values of [tex]C[/tex] is [tex]31[/tex] and [tex]41[/tex].
