# Math Help - Factor/Remainder theorem

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

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

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 $ax$ in the line that I've indicated. When you divide $4x^2-7x-15$ by $x-3$, the result is simply $4x+5$.

If I may expand on acc100jt's solution a little:

If $4x^2-7x-15=(x-3)(Ax+B)$, then when we multiply out the RHS we get:
$4x^2-7x-15=Ax^2+(B-3A)x -3B$
So, when we compare coefficients:
the coefficient of $x^2:4 = A$
and
the constant term: $-15 = -3B \Rightarrow B = 5$
Check with the coefficient of $x$:
$B-3A = 5 -12 = -7$, which is correct
So the other factor is $Ax+B = 4x+5$

Hello s2951

I'm not sure where you get the term $ax$ in the line that I've indicated. When you divide $4x^2-7x-15$ by $x-3$, the result is simply $4x+5$.

If I may expand on acc100jt's solution a little:

If $4x^2-7x-15=(x-3)(Ax+B)$, then when we multiply out the RHS we get:
$4x^2-7x-15=Ax^2+(B-3A)x -3B$
So, when we compare coefficients:
the coefficient of $x^2:4 = A$
and
the constant term: $-15 = -3B \Rightarrow B = 5$
Check with the coefficient of $x$:
$B-3A = 5 -12 = -7$, which is correct
So the other factor is $Ax+B = 4x+5$

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

6. Here's another way,since $(x-3)$ is a factor, $-3$ is a root.
We can do the following,
$f(x)-f(-3)= 4x^2-7x-15-4(-3)^{2}+7\times3+15$ $=4(x^2-(-3)^2)-7(x-1)=4(x+3)(x-3)-7(x-3)=(x-3)(4x+5)$

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

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

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

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

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

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

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

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