Question

Practice Task 2: Factor if possible. If polynomial is not factorable, then write PRIME. 1. x2 + 9x + 14 2. y2 + 11y + 24 3. Z2 - 6z + 8 4. m2 + 2m - 35 5. n2 - 11n - 42​

Answer 1

Step-by-step explanation:

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Answer 2

Given:

1. $x^{2} +9x+14$

2. $y^{2}+11y+24$

3. $z^{2}-6z+8$

4. $m^{2} +2m-35$

5. $n^{2} -11n-42$

To find:

Factors, if possible. Otherwise, write as PRIME.

Solution:

1. Here, $a=1,b=9,c=14$

$product =14$

$sum=9$

Determine the factors of 14 that give a sum of 9.

Factors of 14 = $(1,14),(2,7)$

Factors of 14 that give sum of 9 is $(2,7)$

$x^{2} +9x+14$

$=x^{2} +2x+7x+14$

$=(x^{2} +2x)+(7x+14)$

Take out the common factors from each parenthesis such that the first factor should be same in second too.

$=x(x+2)+7(x+2)$

Here, $(x+2)$ was remaining after taking out $x$ as common factor in the first parenthesis. Make sure that $(x+2)$ is the common factor in the second parenthesis so that further factorisation will be easier.

Since, $(x+2)$ is the common factor, combine the other two to form the second factor.

$=(x+2)(x+7)$

∴ $x^{2} +9x+14=(x+2)(x+7)$

2. Here, $Product=24,sum=11$

Factors of 24 = $(1,24),(2,12),(3,8),(4,6)$

Factors of 24 that give a sum of 11 = $(3,8)$

$y^{2}+11y+24$

$=y^{2}+8y+3y+24$

$=(y^{2}+8y)+(3y+24)$

Take out $y$ and $3$ as common factors from first and second parentheses respectively.

$= y(y+8)+3(y+8)$

Take out $(y+8)$ as common factor.

$=(y+8)(y+3)$

∴ $y^{2}+11y+24$ $=(y+8)(y+3)$

3. $product=8,sum=-6$

Since, product is positive and sum is negative, both the factors have to be negative.

Factors of 8 = $(-1,-8),(-2,-4)$

Factors of 8 which gives sum as $-6=(-4,-2)$

$z^{2}-6z+8$

$= z^{2}-4z-2z+8$

$=(z^{2}-4z)-(2z-8)$

Taking out $z$ and $2$ as common factors from first and second parentheses respectively.

$=z(z-4)-2(z-4)$

Taking out $(z-4)$ as common factor.

$=(z-2)(z-4)$

∴ $z^{2}-6z+8$ $=(z-2)(z-4)$

4. $Product=-35,sum=2$

Product is negative and sum is positive. When a negative is multiplied with positive, product is positive. For sum to be positive, smaller factor is negative while the bigger factor is positive.

Factors of $-35=(35,-1),(7,-5)$

Factors of $-35$ having sum as $2=(7,-5)$

$m^{2} +2m-35$

$= m^{2}+7m-5m-35$

$=(m^{2} +7m)-(5m+35)$

Factor out $m$ and $5$ from first and sencond parenthesis respectively.

$=m(m+7)-5(m+7)$

Factor out $(m+7)$

$=(m+7)(m-5)$

∴ $m^{2} +2m-35$ $=(m+7)(m-5)$

5. $Product =-42,sum=-11$

Here, both product and sum are negative. For sum to be negative, bigger factor has to be negative while smaller one has to be positive.

Factors of $-42=(1,-42),(2,-21),(3,-14),(6,-7)$

Factors of $-42$ whose sum is $-11=(3,-14)$

$n^{2} -11n-42$

$=n^{2}-14n+3n-42$

$=(n^{2}-14n)+(3n-42)$

Factor out $n$ and $3$ from the first and second parentheses respectively.

$=n(n-14)+3(n-14)$

Factor out $(n-14)$

$=(n-14)(n+3)$

∴ $n^{2} -11n-42$ $=(n-14)(n+3)$

The following are the factors for these polynomials:

1. $x^{2} +9x+14=(x+2)(x+7)$

2. $y^{2}+11y+24$ $=(y+8)(y+3)$

3. $z^{2}-6z+8$ $=(z-2)(z-4)$

4.  $m^{2} +2m-35$ $=(m+7)(m-5)$

5. $n^{2} -11n-42$ $=(n-14)(n+3)$