1. ## Factor/Remainder theorem

(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

2. Since $\displaystyle (x-3)$ is a factor,
so $\displaystyle f(x)=(x-3)(Ax+B)$
Thus $\displaystyle 4x^{2}-7x-15=(x-3)(Ax+B)$
By comparison, $\displaystyle A=4,B=-5$
Hence, $\displaystyle f(x)=(x-3)(4x+5)$

Anyway, you can get the two factors by the method Cross Multiplication for the quadratic expression, that will be faster.

3. Originally Posted by acc100jt
Since $\displaystyle (x-3)$ is a factor,
so $\displaystyle f(x)=(x-3)(Ax+B)$
Thus $\displaystyle 4x^{2}-7x-15=(x-3)(Ax+B)$
By comparison, $\displaystyle A=4,B=-5$
Hence, $\displaystyle f(x)=(x-3)(4x+5)$

Anyway, you can get the two factors by the method Cross Multiplication for the quadratic expression, that will be faster.
Ok, then how you get (4x+5) ?

4. 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 $\displaystyle Ax+B = 4x+5$

Hello s2951

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 $\displaystyle Ax+B = 4x+5$

Thanks, just solved and its correct. The letter "A" is just symbol to find the value, its redundant.

6. Here's another way,since $\displaystyle (x-3)$ is a factor,$\displaystyle -3$ is a root.
We can do the following,
$\displaystyle f(x)-f(-3)= 4x^2-7x-15-4(-3)^{2}+7\times3+15$$\displaystyle =4(x^2-(-3)^2)-7(x-1)=4(x+3)(x-3)-7(x-3)=(x-3)(4x+5) 7. Hello Raoh Originally Posted by Raoh Here's another way,since \displaystyle (x-3) is a factor,\displaystyle -3 is a root. We can do the following, \displaystyle f(x)-f(-3)= 4x^2-7x-15-4(-3)^{2}+7\times3+15$$\displaystyle =4(x^2-(-3)^2)-7(x-1)=4(x+3)(x-3)-7(x-3)=(x-3)(4x+5)$
I think what you mean is:
Since $\displaystyle (x-3)$ is a factor, $\displaystyle +3$ is a root $\displaystyle \Rightarrow f(3)=0$
So:
$\displaystyle f(x) = f(x) - f(3) = 4(x^2-3^2) - 7(x-3) -(15-15)$
$\displaystyle =4(x+3)(x-3) -7(x-3)$

$\displaystyle =(4x+12)(x-3) -7(x-3)$

$\displaystyle =(4x+12-7)(x-3)$

$\displaystyle =(4x+5)(x-3)$

Hello RaohI think what you mean is:
Since $\displaystyle (x-3)$ is a factor, $\displaystyle +3$ is a root $\displaystyle \Rightarrow f(3)=0$
So:
$\displaystyle f(x) = f(x) - f(3) = 4(x^2-3^2) - 7(x-3) -(15-15)$
$\displaystyle =4(x+3)(x-3) -7(x-3)$

$\displaystyle =(4x+12)(x-3) -7(x-3)$

$\displaystyle =(4x+12-7)(x-3)$

$\displaystyle =(4x+5)(x-3)$