Multiplication is often expressed as repeated addition.
Such as
$$5\cdot 3=5+5+5$$ $$-5\cdot 3=(-5)+(-5)+(-5)$$
Above in both the cases multiplier is positive.In case of multiplier is negative how will you express following as repeated addition
$$5\cdot (-3)$$
and
$$-5\cdot (-3)?$$
Do can they expressed as repeated subtraction from zero as follows
$$5\cdot (-3)=0-(5)-(5)-(5)$$
$$-5\cdot (-3)=0-(-5)-(-5)-(-5)$$
On
First a correction -- it's the multiplier or scalar that's written first. Thus, the multiplier in the expression $ax$ is the $a,$ which scales or multiplies the $x.$
Now to your question about integer multiplication, in particular those involving negative integers, we should first clear ourselves what we mean by a negative number. First, the unit $-1,$ what is this? Well, this is simply a new number (with respect to the nonnegative integers) with the property that when added to $1$ gives $0.$ That's what the sign $-$ preceding the $1$ tells us. Thus, it's the opposite of the other unit (it's often called its negative). Then we may scale up this negative number to any length to get the opposite of every other positive number. Thus, $3×(-1)=-3,$ namely that which added to $3$ gives nothing. Similarly for other positive numbers.
Now we have seen how to multiply any number by a positive number (and zero). What does it mean to multiply a number by a negative number? Well, again, what would it mean to have $(-1)×8,$ say? Recall that $1×8=8.$ That is, multiplying by the positive unit leaves the multiplicand unchanged. So what happens when we multiply by the negative unit? We figure this out by looking at the product $(-1)×1.$
What could $-1×1$ mean? If we want to keep the commutative property (and we do!) then we may require that $-1×1=1×-1.$ If this is the case, then it follows that $$-1×1=-1.$$ This leads us to the fundamental property of the unit $-1$ in multiplication -- it changes the sign of the multiplicand to the opposite, from $+$ to $-$ and from $-$ to $+.$ It now follows that $-1×-1$ must be $1.$ Another way to see this is to require that the distributive rule continue to hold, so that we have $$(-1+1)^2=(-1)(-1)+(-1)(1)+(1)(-1)+(1)(1)=(-1)(-1)-1-1+1=(-1)(-1)-1,$$ on the one hand, and on the other $$(-1+1)^2=0^2=0,$$ so that we see that $(-1)(-1)-1=0,$ or as before that $$(-1)(-1)=1.$$
Yet another way to see this is to extend the concept of negativity to numbers of the form $-(\text{positive number})$ themselves, as nothing stops this. Anyway, we soon discover this loops back on itself, as the negative of $-1,$ for example is $1.$ Why? Because the number which when added to $-1$ gives $0$ is nothing but $1.$ Hence we have shown that $-(-1)=1.$ Since $-1$ means $-1×1,$ and in general $-a=-1×a,$ it follows again that $$-(-1)=-1×(-1)=1.$$
I think
$$ 5\cdot (-3) $$
can be read as "five times minus three" and can be expressed as $$ (-3)+(-3)+(-3)+(-3)+(-3)=-3-3-3-3-3=-15 $$
But when both factors are negative then the "repeated addition" is sort of losing its meaning. You could go by doing as follows instead
$$ (-5)\cdot (-3)=(-5)\cdot(-1)\cdot 3= $$ since multiplication is commutative you can rearrange the terms
$$ =(-1)\cdot (-5)\cdot3= $$ since multiplication is also associative you can start with the second multiplication
$$ =(-1)\cdot [(-5)+(-5)+(-5)]=-1\cdot (-15)= $$
and here you could argue that multiplying with $-1$ changes sign so you end up with
$$ =15. $$