Question

Factor, if possible. If the polynomial is prime, say so

Answer 1

Factoring first expression

$w^{2} +8w +16$

we can split $w^{2}$ into two parentheses like this

(w )(w )------------------------------------------------(1)

Now to fill the blanks in each parentheses we make factors of constant term 16 and will pick those factors of 16 whose sum is middle term 8

So we can see pairs of factors of 16 as

(1,16); (2,8); (4,4).

out of these we can see that 4+4 gives 8. So +4 and +4 are pairs of factors of 16 that will work here.

So simply plug +4 in blank in first parentheses and +4 in blank in other parentheses in (1)

(w +4)(w +4)

so that the factored form.

We can also write this as $(w+4)^{2}$

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To factor $35x^{2} -2x -1$ we will use AC method.

In this method we will first compare this expression with standard form

$ax^{2} +bx +c$

$35x^{2} -2x -1$

so a = 35, b = -2, c = -1

So ac = 35 × -1 = -35

So we will make factors of ac which is -35 here and will pick those factors of -35 which will give sum as b which is -2.

So pairs of factors of 35 are

(1,35); (5,7)

We can see that we can get sum as -2 form 5 and 7 if we take +5 and -7 as +5-7 =-2. So +5 and -7 factors will work here

So now split the middle term -2x in $35x^{2} -2x -1$ in factors +5 and -7. So we can rewrite -2x as +5x-7x

$35x^2 +5x -7x -1$

Now we will make two groups as shown

$(35x^2 +5x)+( -7x -1 )$

Now factor each group separately

in $(35x^2 +5x)$ we have greatest common factor as 5x so take that common out and divide all terms by 5x to get the terms left inside parentheses.

So we get 5x(7x+1) for this group

Similarly to factor (-7x-1) we can take greatest common factor -1 common out so we will have -1(7x+1)

5x(7x+1) -1(7x+1)

Now both terms have (7x+1) factor common so take that common out from both terms. So we will have

(7x+1)(5x-1)

So thats its factored form

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Again for third expression $3y^{2} + 10y + 7$

we will use AC method to factor

So here a = 3, b = 10, c = 7

so ac = 3 × 7 = 21

So we will make factors of 21 and will pick those factors of 21 whose sum is b which is 10

Pair factors of 21 are

(1,21); (3,7)

We can see that factors 3+7 can give 10. So factors +3 and +7 will work here.

so split the middle term 10y in $3y^{2} + 10y + 7$ in terms of factors +3 and +7 so we can rewrite 10y as +3y+7y

$3y^{2} + 3y + 7y+ 7$

Now again make groups

$(3y^{2} + 3y ) + (7y+ 7)$

Factor each group separately by taking greatest common factor out in each group

3y(y+1) + 7(y+1)

As we can see both terms have factor (y+1) common in them so we can take (y+1) factor common out. So we will have final factored form as

(y+1)(3y+7)

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