Question

How to split the middle term of a polynomial 6x^2-3-7x?

Answer 1

$p(x)=6x^2-7x-3=0 \\ \\ \\ a=6\ \ \ ; \ \ \ b=-7\ \ \ ; \ \ \ c=-3$

Here, the middle term is bx, i.e., -7x. Let b=-7 be spit as p+q.

For a quadratic polynomial ax² + bx + c = 0, if the middle term bx is split as p+q, then pq = ac.

Here,

$p+q=-7=b\ \ \ \ \ \longrightarrow\ \ \ (1) \\ \\ \\ pq=6 \times -3 \\ \\ pq=-18=ac \\ \\ \\ p-q=\sqrt{b^2-4ac} \\ \\ p-q=\sqrt{(-7)^2-(4 \times -18)} \\ \\ p-q=\sqrt{49-(-72)} \\ \\ p-q=\sqrt{121} \\ \\ p-q=\pm 11\ \ \ \ \ \longrightarrow\ \ \ (2)$

Whenever we take p - q as either 11 or -11, we get same values for p and q as interchanged. Let me take p - q = 11.

$(1)+(2) \\ \\ (p+q)+(p-q)=-7+11 \\ \\ 2p=4 \\ \\ p=2 \\ \\ \\ (1)-(2) \\ \\ (p+q)-(p-q)=-7-11 \\ \\ 2q=-18 \\ \\ q=-9$

$\therefore\ \bold{-7x=2x-9x\ \ \ \ \boxed{\bold{OR}}\ \ \ \ -7x=-9x+2x} \\ \\ \\ \\ 6x^2-7x-3 \\ \\ \\ 6x^2+2x-9x-3\ \ \ \ \ \ \ \ \ \ \ \ \ \ \boxed{\bold{OR}}\ \ \ \ \ \ \ \ \ \ \ \ \ \ 6x^2-9x+2x-3 \\ \\ 2x(3x+1)-3(3x+1)\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 3x(2x-3)+1(2x-3) \\ \\ \\ (2x-3)(3x+1) \\ \\ x=\frac{3}{2}\ \ \ ; \ \ \ x=-\frac{1}{3}$

Hope this helps you. ^_^

Please mark it as the brainliest if this helps you.

Please ask me if you have any doubts.

Thank you. :-))