Hello s2951 Originally Posted by
s2951 Ok, then how you get (4x+5) ?
Originally Posted by
s2951 (x-3) is the Factor of f(x) 4x^2-7x-15
Help me is this correct of finding the other factor?
f(x)=l(x)q(x)+R
f(x)=(x-3)(4x+ax+5)+0
f(x)=x(4x+ax+5)-3(4x+ax+5)
f(x)=4x^2+ax^2+5x-12x-3ax-15
f(x)=(ax+5)=0 (12-3ax)=0
ax=-5 ax=4
I'm not sure where you get the term $\displaystyle ax$ in the line that I've indicated. When you divide $\displaystyle 4x^2-7x-15$ by $\displaystyle x-3$, the result is simply $\displaystyle 4x+5$.
If I may expand on acc100jt's solution a little:
If $\displaystyle 4x^2-7x-15=(x-3)(Ax+B)$, then when we multiply out the RHS we get:
$\displaystyle 4x^2-7x-15=Ax^2+(B-3A)x -3B$
So, when we compare coefficients:
the coefficient of $\displaystyle x^2:4 = A$
and
the constant term: $\displaystyle -15 = -3B \Rightarrow B = 5$
Check with the coefficient of $\displaystyle x$:
$\displaystyle B-3A = 5 -12 = -7$, which is correct
So the other factor is $\displaystyle Ax+B = 4x+5$
Grandad