\(\displaystyle x\)-intercepts occur when \(\displaystyle y=0\).

1. Set the expression equal to \(\displaystyle 0\) and rearrange:

\(\displaystyle x^{2}-10x+9=0\)

2. Factor the expression:

\(\displaystyle (x-1)(x-9)=0\)

3. Solve for \(\displaystyle x\):

\(\displaystyle x-1=0\)

\(\displaystyle x=1\)

and...

\(\displaystyle x-9=0\)

\(\displaystyle x=9\)

4. Rewrite the answers as coordinates:

\(\displaystyle x=9\) becomes \(\displaystyle (9,0)\) and \(\displaystyle x=1\) becomes \(\displaystyle (1,0)\).

1. Factor the expression:

\(\displaystyle 6x^{2}+2x-20= (3x-5)(2x+4)\)

2. Solve for \(\displaystyle x\):

\(\displaystyle 3x-5=0\)

\(\displaystyle x=\frac{5}{3}\)

and...

\(\displaystyle 2x+4=0\)

\(\displaystyle x=-2\)

Remember that when you factor a trinomial in the form \(\displaystyle ax^2+bx+c\), you need to find factors of \(\displaystyle c\) that add up to \(\displaystyle b\).

First, write down all the possible factors of \(\displaystyle c\).

\(\displaystyle 1, 15\)

\(\displaystyle -1, -15\)

\(\displaystyle 3, 5\)

\(\displaystyle -3, -5\)

Then add them to see which one gives you the value of \(\displaystyle b\)

\(\displaystyle 1+15=16\)

\(\displaystyle -1-15=-16\)

\(\displaystyle 3+5=8\)

\(\displaystyle -3-5=-8\)

Thus, the factored form of the expression is \(\displaystyle (x-3)(x-5)\)

First, find any common factors. In this case, there is a common factor: \(\displaystyle 9x^8\)

\(\displaystyle 9x^8(x^2+2x-3)\)

Now, factor the trinomial.

To factor the trinomial, you will need to find factors of \(\displaystyle -3\) that add up to \(\displaystyle 2\).

List out the factors of \(\displaystyle -3\), then add them.

\(\displaystyle -1+3=2\)

\(\displaystyle 1-3=-2\)

Thus, \(\displaystyle x^2+2x-3=(x-1)(x+3)\)

This question calls for us to factor the polynomial into two binomials. Since the first term is \(\displaystyle x^2\) and the last term is a number without a variable, we know that how answer will be of the form \(\displaystyle (x+ a)(x+ b)\) where a and b are positive or negative numbers.

To find a and b we look at the second and third term. Since the second term is \(\displaystyle 9x\) we know \(\displaystyle a+b=9\). (The x comes from a and b multiplying by x and then adding with each other). The +10 term tells us that \(\displaystyle a*b=10\). Using these two pieces of information we can look at possible values. The third term tells us that 1 & -10 and -1 & 10 are the possible pairs. Now we can look and see which one adds up to make 9. This gives us the pair -1 & 10 and we plug that into the equation as a and b to get our final answer.

\(\displaystyle (x+10)(x-1)\)

This question calls for us to factor the polynomial into two binomials. Since the first term is \(\displaystyle x^2\) and the last term is a number without a variable, we know that how answer will be of the form \(\displaystyle (x+ a)(x+ b)\) where a and b are positive or negative numbers.

To find a and b we look at the second and third term. Since the second term is \(\displaystyle 5x\) we know \(\displaystyle a+b=5\). (The x comes from a and b multiplying by x and then adding with each other). The -14 term tells us that \(\displaystyle a*b=-14\). Using these two pieces of information we can look at possible values. The third term tells us that 1 & -14, 2 & -7, -2 & 7, and -1 & 14 are the possible pairs. Now we can look and see which one adds up to make 5. This gives us the pair -2 & 7 and we plug that into the equation as a and b to get our final answer.

\(\displaystyle (x+7)(x-2)\)

This question calls for us to factor the polynomial into two binomials. Since the first term is \(\displaystyle x^2\) and the last term is a number without a variable, we know that how answer will be of the form \(\displaystyle (x+ a)(x+ b)\) where a and b are positive or negative numbers.

To find a and b we look at the second and third term. Since the second term is \(\displaystyle -9x\) we know \(\displaystyle a+b=-9\). (The x comes from a and b multiplying by x and then adding with each other). The \(\displaystyle +8\) term tells us that \(\displaystyle a*b=8\). Using these two pieces of information we can look at possible values. The third term tells us that 1 & 8, 2 & 4, -2 & -4, and -1 & -8 are the possible pairs. Now we can look and see which one adds up to make -9. This gives us the pair -1 & -8 and we plug that into the equation as a and b to get our final answer.

\(\displaystyle (x-1)(x-8)\)

To multiply trinomials, simply foil out your factored terms by multiplying each term in one trinomial to each term in the other trinomial. I will show this below by spliting up the first trinomial into its 3 separate terms and multiplying each by the second trinomial.

\(\displaystyle x^2(2x^2+4x-6)-3x(2x^2+4x-6)+1(2x^2+4x-6)\)

Now we treat this as the addition of three monomials multiplied by a trinomial.

\(\displaystyle 2x^4+4x^3-6x^2-6x^3-12x^2+18x+2x^2+4x-6\)

Now combine like terms and order by degree, largest to smallest.

\(\displaystyle 2x^4-2x^3-16x^2+22x-6\)

The \(\displaystyle 5\) is distributed and multiplied to each term \(\displaystyle x\), \(\displaystyle y\), and \(\displaystyle z\).

\(\displaystyle 7\) is multiplied to both \(\displaystyle x\) and \(\displaystyle y\) and \(\displaystyle 8\) is only multiplied to \(\displaystyle z\).