Mc Graw Hill Key To Algebra Book 4 Polynomials 1st Edition Chapter 8 Multiplying And Factoring Polynomials
Page 16 Exercise 1 Answer
Given the product expression: (x+3)(x+2)
To multiply out the two binomials x+3 and x+2 and write its product.
Now we multiply out each term of one polynomial to another polynomial:
Given the product expression:(x+3)(x+2)
(x+3)(x+2)=x(x+2)+3(x+2)
(using the distributive property)
=x2+5x+3.2
(sincex1.x2=x1+1 and combining like terms of the power of x
=x2+5x+6
The product we can write:
3.2=3+3
=6
The final multiplied expression is: (x+3)(x+2) = x2 + 5x + 6
Page 16 Exercise 4 Answer
Given the product expression: (y+4)(y+10)
Multiply the binomials (y+4) and (y+10) and write its product.
Now we multiply out each term of one polynomial to another polynomial:
Given the product expression(x+3(x+2)
(x+3)(x+2)= x
The final multiplied expression is: (y+4)(y+10) = y2 + 14y + 40
Page 16 Exercise 5 Answer
Given the product expression: (x+5)(x+6)
Multiply the binomials (x+5) and (x+6) and write the product.
Now we multiply out each term of one polynomial to another polynomial:
(x+5)(x+6)=x(x+6)+5(x+6)
=x.x+6x+5x+5.6 (using distributive law)
=x2+11x+5.6
(since x1.x1=x1+1 and combining like terms of x
The= x2+11x+30
The product can be viewed as
5.6=6+6+6+6+6
=12+12+6
=30
The final multiplied expression is: (x+5)(x+6) = x2 + 11x + 30
Page 16 Exercise 6 Answer
Given the product expression: (x+1)(x+6)
To multiply the binomials (x+5) and (x+6) and write its product.
Now we multiply out each term of one polynomial to another polynomial:
Given the product expression: (y+4)(y+10)
(y+4)(y+10)=y(y+10)4(y+10)
=y.y+10y+4y+4.10
(using distributive law)
=y2+14y+4.10
(since x1.x1= x1+1 and combining like terms of power of x
=y2+14y+4.10
(since x1.x1=x1+1 and combining like terms of power of x)
=y2+14y +40
The product can be viewed as:
4.10=10+10+10+10
=20+20
=40
The final multiplied expression is: (x+1)(x+6) = x2 + 7x + 6
Page 16 Exercise 7 Answer
Given product expression: (a+8)(a+9)
Multiply the binomials (a+8) and (a+9) and write its product.
Now we multiply out each term of one polynomial to another polynomial:
Given the product expression:(x+5)(x+6)
(x+5)(x+6)=x(x+6)+5(x+6)
=x.x+6x+5x+5.6 (using distributive law)
=x2+11x+5.6
(since x1.x1=x1+1 and combining like terms of x
The= x2+11x+30
The product can be viewed as
5.6=6+6+6+6+6
=12+12+6
=30
The final multiplied expression is: (a+8)(a+9) = a2 + 17a + 72
Page 16 Exercise 8 Answer
Given product expression: (z+3)(z+3)
To multiply the binomials (z+3) and (z+3) and write its product.
Now we multiply out each term of one polynomial to another polynomial:
Given product expression: (z+3)(z+3)
(z+3)(z+3)=z(z+3)+3(z+3)
=z.z+3z+3z+3.3
(using distributive law)
=za²+6z+9
(since z¹.z¹=z¹+1 combing like terms of z)
=z²+6.z+9
The product can be viewed as
3.3=3+3+3
=6+3
=9
The final multiplied expression is: (z+3)(z+3) = z2 + 6z + 9
Page 16 Exercise 1 Answer
Given the product expression: (x−4)(x−5)
Multiply the binomials (x−4) and (x−5) and find the product value.
Now we multiply out each term of one polynomial to another polynomial:
Given that product expression:(x-4)(x-5)
(x-4)(x-5)=x(x-5)-4(x-5)
=x.x-5x-4x+(-4).(-5)
(using distributive law)
=x2-9x+(-4)++.(-5)(combing like terms)
=x2-9x+20
The product can be viewed as:
(-4).(-5)=4.5
=5+5+5+5
=10+10
=20
The expanded form of the product can be viewed as (x−4)(x−5) = x2 − 9x + 20
Page 16 Exercise 2 Answer
Given product expression : (x−2)(x−4)
Multiply the binomials (x−4) and (x−2) and write their product value.
Now we multiply out each term of one polynomial to another polynomial:
Given Product expression: (x-2)(x-4)
(x-2)(x-4)=x(x-4)-2(x-4)
=x.x-4x-2x+(-2).(-4)
(using distributive law)
=x²-6x+(-2).(-4)
(combining like terms)
=x2-6x+8
the product can be viewed as
(-2).(4)=2.4
=4+4
=8
The expanded form of the product is: (x−2)(x−4) = x2 − 6x + 8
Page 16 Exercise 3 Answer
Given the product expression: (x−3)(x−5)
Multiply the binomials (x−3) and (x−5) and write the value of the product.
Now we multiply out each term of one polynomial to another polynomial:
Given Product Expession: (x-3)(x-5)
(x-3)(x-5)=x(x-5)-3(x-5)
=x.x-5x-3x+(-3).(-5)
(using distributive law)
= x2-8x+(-3).(-5)
(combining like terms)
=x2-8x+15
the product can be viewed as:
(-3).(-5)=3.5
=5+5+5
=10+5
=15
The expanded form of the product is: (x−3)(x−5) = x2 − 8x + 15
Page 16 Exercise 4 Answer
Given product expression: (x−6)(x−6)
Multiply the binomials (x−6) and (x−6) and write the product value.
Now we multiply out each term of one polynomial to another polynomial:
Given product expression:(x-6)(x-6)
(x-6)(x-6)=x(x–6)-6(x-6)
=x.x-6x-6x+(-6).(-6)
(using distributive law)
=x2-12x+(-6).(-6)
(combing like terms)
=x2-12x+36
the product can be viewed as:
(-6).(-6)=6.6
=6+6+6+6+6+6
=12+12+12
=24+12
=36
The expanded form of the product is: (x−6)(x−6) = x2 − 12x + 36
Page 16 Exercise 5 Answer
Given the product expression: (x−1)(x−8)
Multiply the binomials (x−1) and (x−8) and write the product value.
Now we multiply out each term of one polynomial to another polynomial:
Given product expression: (x-1)(x-8)
(x-1)(x-8)= x(x-8)-1(x-8)
=x.x-8x-x+(-1).(x-8)
(using distributive law)
=x2-9x+8
(combining like terms)
the product can be viewed as:
(-1).(-8)=8.1
=8
= 8 (Since the product of any number with 1 is the number itself)
The expanded form of the product is: (x−1)(x−8) = x2 − 9x + 8
Page 16 Exercise 6 Answer
Given the product expression: (x−7)(x+3)
Multiply the binomials (x−7) and (x+3) and write the product value.
Now we multiply out each term of one polynomial to another polynomial:
Given:(x-7)(x+3)
(x-7)(x+3)=x(x+3)-7(x+3)
=x.x+3x-7x+3.(-7)
(using the distributive law)
=x2-4x+3.(-7)
(combining like terms)
=x2-4x-21
The product can be viewed as:
3.(-7)=-7-7-7
=-14-7
=-21
The expanded form of the product is : (x−7)(x+3) = x2 − 4x − 21
Page 16 Exercise 7 Answer
Given the product expression: (x−5)(x+3)
Multiply the binomials (x−5) and (x+3) and write its product value.
Now we multiply out each term of one polynomial to another polynomial:
Given: (x-5)(x+3)
(x-5)(x+3)=x(x+3)-5(x+3)
=x.x+3x-5x+(-5).3
(using distributive law)
=x2-2x+(-5).3
(combining like terms)
=x2-2x-15
the product can be viewed as:
3.(-5)=-5-5-5
=-10-5
=-15
The expanded product can be written as: (x−5)(x+3) = x2 − 2x − 15
Page 16 Exercise 8 Answer
Given product expression: (x+2)(x−6)
Multiply the binomials (x+2) and (x−6) and write its product value.
Now we multiply out each term of one polynomial to another polynomial:
Given product expression: (x+2)(x-6)
(x+2)(x-6)=x(x-6)+2(x-6)
=x.x-6x+2x+(2).(-6)
(using the distributive property)
= x2-4x+2.(-6)
The product can be viewed as:
2.(-6)=-6-6
=-12
The expanded form of the product: (x+2)(x−6) = x2 − 4x − 12
Page 16 Exercise 9 Answer
Given product expression: (x+8)(x−5)
Multiply the binomials (x+8) and (x−5) and write its value.
Now we multiply out each term of one polynomial to another polynomial:
Given Product expression:(x+8)(x-5)
(x+8)(x-5)=x(x-5)+8(x-5)
=x.x-5x+8x+8.(-5)
(using the distributive property)
=x2-3+8.(-5)
(combining like terms)
=x2-3x-40
The product can be viewed as:
8.(-5)=-5-5-5-5-5-5-5-5
=-10-10-10-10
=-20-20
=-40
The product in the expanded form is: (x+8)(x−5) = x2 + 3x − 40
Page 16 Exercise 10 Answer
Given product expression: (x−8)(x+5)
Multiply the binomials (x−8) and (x+5) and write its value.
Now we multiply out each term of one polynomial to another polynomial:
Given product expression: (x-8)(x+5)
The expanded form of the product is: (x−8)(x+5) = x2 − 3x − 40
Page 17 Exercise 1 Answer
Given product expression: (x−3)(x−2)
To multiply the binomials (x−2) and (x−3) and write the value.
Now we multiply out each term of one polynomial to another polynomial:
Given product expression: (x-3)(x-2)
(x-3)(X-2)=x(x-2)-3(x-2)
=x.x-2x-3x+(-3).(-2)
=x.x-2x-3x+(-3).(-2)
(using distributive property)
=x2-5x+(-3).(-2)
(combining like terms)
=x2-5x+6
The product can be viewed as
(-3).(-2)=3.2
=3+3
=6
The expanded form of the product is: (x−3)(x−2) = x2 − 5x + 6
Page 17 Exercise 2 Answer
Given product expression: (x−1)(x−1)
To multiply the binomials (x−1) and (x−1) and write the value.
Now we multiply out each term of one polynomial to another polynomial:
Given product expression: (x-1)(x-1)
(x-1)(x-1)=x(x-1)-1(x-1)
x.x-x-x+(-1)(-1)
(using the distributive property)
=x2-2x+1
(combining like terms)
The product can be viewed as:
(-1).(-1)=1.1
=1
The expanded form of the product is: (x−1)(x−1) = x2 − 2x + 1
Page 17 Exercise 3 Answer
Given the product expression: (x−7)(x−2)
To multiply the binomials (x−7) and (x−2) and write its value.
Now we multiply out each term of one polynomial to another polynomial:
Given the product expression:(x-7)(x-2)
(X-7(x-2)=x(x-2)-7(x-2)
=x.x-2x-7x+(-7)(-2)
(using distributive property)
=x²-9x+(-7)(-2)
(combining like terms)
=x²-9x+14
The product can be viewed as:
(-7).(-2)=7.2
=7+7
=14
The expanded form of the product is: (x−7)(x−2) = x2 − 9x + 14
Page 17 Exercise 4 Answer
Given the product expression: (x−3)(x−4)
To multiply the binomials (x−3) and (x−4) and write its value.
Now we multiply out each term of one polynomial to another polynomial:
Given the product expression: (x-3)(x-4)
(x-3)(x-4)=x(x-4)-3(x-4)
=x.x-4x-3x+(-3).(-4)
(using distributive law)
=x2-7x+12(combining like terms)
The product can be viewed as:
(-3).(-4)=3.4
=4+4+4
=8+4
=12
The expanded form of the product is: (x−3)(x−4) = x2 − 7x + 12
Page 17 Exercise 5 Answer
Given the product expression: (x+8)(x−3)
Multiply the binomials (x+8) and (x−3) and write its value.
Now we multiply out each term of one polynomial to another polynomial:
Given the product expression:(x+8)(x-3)
(x+8)(x-3)=x(x-3)+8(x-3)
=x.x-3x+8x+8(-3)
(using distributive law)
=x²+5x-24(combining like terms)
The product can be viewed as:
8.(-3)=-3-3-3-3-3-3-3-3
=-6-6-6-6
=-12-12
=-24
The expanded form of the product is: (x+8)(x−3) = x2 + 5x − 24
Page 17 Exercise 6 Answer
Given product expression: (x+5)(x−2)
To multiply the binomials (x+5) and (x−2) and write its value.
Now we multiply out each term of one polynomial to another polynomial:
Given product expression:(x+5)(x-2)
(x+5)(x-2)=x(x-2)+(x-2)
=x.x-2x+5x+5.(-2)
(using distributive law)
=x2+3x-10 (combining like terms)
The product can be viewed as:
5.(-2)=-2-2-2-2-2
-4-4-2
The expanded form of the product is: (x+5)(x−2) = x2 + 3x − 10
Page 17 Exercise 7 Answer
We have given:
(a−4)(a−6)
We have to find the above product.
We will multiply the above two expressions to find the resultant polynomial.
Given rectangle length =x+4
Breadth=x-2
A=lxw
we have, l=x+4
w=x-2
The area of a rectangle is A=lxw
A & =(x+4)(x-2) \\
& =x^2-2 x+4 x-8 \\
& =x^2+2 x-8
\end{aligned}\)
This is a quadratic polynomial.
The multiplication (a−4)(a−6) is equal to a2 − 10a + 24.
Page 17 Exercise 10 Answer
We have given:
(x+4)(x−4)
We have to find the above product.
We will use the identity
(a+b)(a−b) = a2 − b2 to find the resultant polynomial.
Given: (a-4)(a-6)
multiply the expressions
(a-4)(a-6)
multiply the expressions
(a-4)(a-6)
=a(a-6)-4(a-6)
=a(a-6)-4(a-6)
=a2-6a-4a+24
=a2-10a+24
This is the required quadratic polynomial.
The product(x+4)(x−4) is equal to x2 − 16.
Page 17 Exercise 12 Answer
We have given:
(x−3)(x+4)
We have to find the above product.
We will multiply the above two expressions to find the resultant polynomial.
Given:(x-3)(x+4)
multiply the expressions
(x-3)(x+4)
=x(x+4)-3(x+4)
=x²+4x-3x-12
=x²+x-12
This is the required polynomial.
The product(x−3)(x+4) is equal to x2 + x − 12.
Page 17 Exercise 14 Answer
We have given:
(x−6)(x+6)
We have to find the above product.
We will use the identity(a−b)(a+b) = a2 − b2 here and find the resultant polynomial.
Given:(x-6)(x+6)
multiply the expressions:
(x-6)(x+6)
=x²-6²
(since(x-a)(x+a)=x2-a2]
=x²-36
This is the resultant polynomial.
The product(x−6)(x+6) is equal to x2 − 36.
Page 17 Exercise 15 Answer
We have given:
(y+3)(y−3)
We have to find the above product.
We will multiply the above two expressions to find the resultant polynomial.
Given:(y+3)(y-3)
⇒ multiply the expressions
(y+3)(y-3)
=y(y-3)+3(y-3)
=y²-3y+3y-9
=y²-9
This is the resultant quadratic polynomial.
The product (y+3)(y−3) is equal to y2 − 9.
Page 17 Exercise 16 Answer
Given:
(x+9)(x−2)
We have to find the above product.
We will multiply the above two expressions to find the resultant polynomial.
Given: (x+9)(X-2)
Multiply the expressions
(x+9)(x-2)
=x(x-2)+9(x-2)
=x²-2x+9x-18
=x²+7x-18
This is the required polynomial.
The product(x+9)(x−2) gives a quadratic polynomial x2 + 7x − 18.
Page 17 Exercise 1 Answer
We have given a binomial:
(x+5)
We have to calculate(x+5)2.
We will calculate the resultant polynomial by using the identity
(a+b)2 = a2 + 2ab + b2
Given binomial:(x+5)2
(x+5)2
Here, a=x,b=5
∴(x+5)2=x2+2(x)(5)+4=52
[since(a+b)²=a²+2ab+b²]
=x2²+10x+25
This is the required polynomial.
(x+5)2 gives a quadratic polynomial x2 + 10x + 25.
Page 17 Exercise 2 Answer
We have given a binomial:
(x+4)
We have to calculate (x+4)2.
We will calculate the resultant polynomial by using the identity:
(a+b)2 = a2 + 2ab + b2
Given binomial: (x+4)2
(x+4)2
Here a=x, b=4
∴ (x+4)²= x²+2(x)(4)+4²
[Since (a+b)²=a²+2ab+b²]
=x²+8x+16
This is the resultant quadratic polynomial.
(x+4)2 gives a quadratic polynomial x2 + 8x + 16.
Page 17 Exercise 3 Answer
We have given a binomial:
x + 10
We have to calculate (x+10)2.
We will calculate the resultant polynomial by using the identity:
(a+b)2 = a2 + 2ab + b2
Given binomial:(x+10)2
here a=x,b=10
∴ (x+10)² =x2+2(x)(10)+100
(since) (a+b)²=a²+2ab+b²]
=x²+20x+100
This is the resultant polynomial.
(x+10)2 gives a quadratic polynomial x2 + 20x + 100.
Page 17 Exercise 4 Answer
We have given a binomial:
x−5
We have to calculate (x−5)2.
We will find the resultant polynomial by using the identity:
(a−b)2 = a2 − 2ab + b2
Given binomial:(x-5)²
(x-5)²
Here a=x,b=5
using the algebraic identity
\(\begin{aligned}(a-b)^2 & =a^2-2 a b+b^2 \\
(x-5)^2 & =x^2-2(x)(5)+25 \\
& =x^2-10 x+25
\end{aligned}\)
This is the required polynomial.
(x−5)2 gives a quadratic polynomial x2 − 10x + 25.
Page 17 Exercise 6 Answer
Given expression is (x−1)2
We have to multiply binomial (x−1) to binomial (x−1)
The result of the multiplication of two binomial is polynomial.
Given binomial:(x-5)2
(x-5)2
Here a=x, b=5
Using the algebraic identity
⇒ \(\begin{aligned}
(a-b)^2 & =a^2-2 a b+b^2 \\
(x-5)^2 & =x^2-2(x)(5)+25 \\
& =x^2-10 x+25
\end{aligned}\)
Page 18 Exercise 1 Answer
We have given a rectangle having length l = x + 4 and width w = x − 2.
We have to find the area A of the rectangle.
A = l × w
Here area is a multiplication of two binomial.
Given rectangle length=x+4
breadth=x-2
A=lxw
we have, l=x+4
The area of the rectangle is a =lxw
⇒ \(\begin{aligned}
A & =(x+4)(x-2) \\
& =x^2-2 x+4 x-8 \\
& =x^2+2 x-8
\end{aligned}\)
The area of the rectangle having length (x+4) and width (x-2) is x2 + 2x – 8.
Page 18 Exercise 2 Answer
We have given a rectangle having length l = x + 7 and width w = x − 4.
We have to find the area A of the rectangle.
A = l × w
Here area is the multiplication of two binomials.
Given A rectangle length l=x+7
breadth w=x-4
Area of rectangle a=lxw
we have, l=x+7
w=x-4
A=(x+7)(x-4)
The area of a rectangle having length x + 7 and width x − 4 is x2 + 3x − 28.
Page 18 Exercise 4 Answer
We have given a rectangle having length l = x + 3 and width w = x + 3.
We have to find the area A of the rectangle.
A = l × w
Here area is the multiplication of two binomials.
Given a rectangle length l=x+3
breadth w=x+3
Area of rectangle A= l×w
we have, l=x+3
w=x+3
A=(x+3)(x+3)
=x²+3x+3x+9
=x²+6x+9
The area of rectangle having length (x+3) and width (x+3) is x2 + 6x + 9
Page 18 Exercise 5 Answer
We have given rectangle having length l = (x+6) and width w = x − 1
We have to find the area A of the rectangle.
A = l × w
To find the area of a rectangle we have to multiply the two binomials.
Given a rectangle having length l=x+6
breadth w=x-1
Area of rectangle A=l×w
we have, l= x+6
w=x-1
A=(x+6)(x-1)
=x²-x+6x-6
=x²+5x-6
The area of a rectangle having length l = x + 6 and width w = (x−1) is x2 + 5x − 6.
Page 19 Exercise 2 Answer
We have been given a trinomial that x2 + 21x + 20 = (x+…)(x+..).
We have to fill in each pair of blanks with the right numbers and we have to check our answers.
We will find the result using the factorization method.
The given Trinomial is, x2+21x+20
Factoring out, we get
⇒ \(\begin{aligned}
x^2+21 x+20 & =x^2+x+20 x+20 \\
& =x(x+1)+20(x+1) \\
& =(x+1)(x+20)
\end{aligned}\)
Now let us check answers by multiplying the binomial
(x+1)(x+20)=x.x+20x+x+20
x2+21x+20
We have factorized and filled in the blanks with the right numbers that are, x2 + 21x + 20 = (x+1)(x+20), and checked our answers
Page 19 Exercise 3 Answer
Given expression is x2 + 12x + 20
We have to factorize the polynomial.
We have, x2 + 12x + 20
The factor of 20 adds up to 12 is 10 and 2.
Given expression: x2+12x+20
Factoring out, we get
⇒ \(\begin{aligned}
x^2+12 x+20 & =x^2+10 x+2 x+20 \\
& =x(x+10)+2(x+10) \\
& =(x+2)(x+10)
\end{aligned}\)
Re check:
By using multiplying the binomial
\(\begin{aligned}(x+2)(x+10) & =x \cdot x+2 x+10 x+20 \\
& =x^2+12 x+20
\end{aligned}\)
Hence it is proved that the factors of polynomials are right.
We have factorized the polynomial as x2 + 12x + 20 = (x+10)(x+2)
We have checked the factors of the polynomial.
Page 19 Exercise 4 Answer
Given expression x2 + 14x + 24
We have to factorize the polynomial.
We have, x2 + 14x + 24
The factor of 24 that add up to 14 is 2, 12
Given expression: x2+14×24
Factoring out, we get
⇒ \(\begin{aligned}
x^2+14 x+24 & =x^2+2 x+12 x+24 \\
& =x(x+2)+12(x+2) \\
& =(x+12)(x+2)
\end{aligned}\)
Re-check the answer:
By multiplying the binomial
⇒ \(\begin{aligned}
(x+12)(x+2) & =x^2+2 x+12 x+24 \\
& =x^2+14 x+24
\end{aligned}\)
Hence the factors of polynomials are right.
We have factorized the polynomial as : x2 + 14x + 24 = (x+2)(x+12)
Page 19 Exercise 5 Answer
Given expression x2 + 9x + 20
We have to factorize the polynomial.
We have , x2 + 9x + 20
The factor of 20 which add up to 9 are 5 and 4.
Given expression: x²+9x+20
Factoring out, we get
\(\begin{aligned}x^2+9 x+20 & =x^2+5 x+4 x+20 \\
& =x(x+5)+4(x+5) \\
& =(x+4)(x+5)
\end{aligned}\)
Rechexking the answer:
by using multiplying the binomial
(x+4)(x+5)x2+4x+-5x+20
=x2+9x+20
Hence the factors of polynomials are right.
We have factorized the polynomial as : x2 + 9x + 20 = (x+4)(x+5)
Page 20 Exercise 4 Answer
Given that the second-degree polynomial is, x2 + 7x + 10
Here we have to factor the given polynomial into a product of two binomials.
Factorization rule :
Group the first two terms together and then the last two terms together.
Factor out a GCF from each separate binomial.
Factor out the common binomial.
Note that if we multiply our answer out, we do get the original polynomial.
Given expression: x2+7x+10
factoring out, we get
x^2+7 x+10 & =x^2+5 x+2 x+10 \\
& =x(x+5)+2(x+5) \\
& =(x+2)(x+5)
\end{aligned}\)
Rechecking the answer:
By multiplying the binomial
(X+2)(x+5)=x²+5x+2x+10
=x²+7x+10
The factorization of the polynomial into a product of two binomials is x2 + 7x + 10 = (x+5)(x+2).
Page 20 Exercise 7 Answer
Given that the second-order degree polynomial is, x2 + 5x + 4
Here we have to factor the given polynomial into a product of two binomials.
Factorization rule :
Group the first two terms together and then the last two terms together.
Factor out a GCF from each separate binomial.
Factor out the common binomial.
Note that if we multiply our answer out, we do get the original polynomial.
Given expression: x2+5x+4
Factoring out, we get
x2+5x+4=x2+x+4x+4
=x(x+1)+4(x+1)
=(x+4)(x+1)
Rechecking the answers:
By multiplying the binomial
(x+4)(x+1)=x2+x+4x+4
=x2+5x+4
The factorization of the polynomial into a product of two binomials is: x2 + 5x + 4 = (x+1)(x+4).
Page 20 Exercise 8 Answer
Given that the second-order degree polynomial is, x2 + 2x + 1
Here we have to factor the given polynomial into a product of two binomials.
Factorization rule :
Group the first two terms together and then the last two terms together.
Factor out a GCF from each separate binomial.
Factor out the common binomial.
Note that if we multiply our answer out, we do get the original polynomial.
Given expression: x2+2x+1
Factoring out, we get
x^2+2 x+1 & =x^2+x+x+1 \\
& =x(x+1)+1(x+1) \\
& =(x+1)(x+1) \\
& \quad \text { or } \\
& =(x+1)^2
\end{aligned}\)
Check:
Now consider the binomial
⇒ \(\begin{aligned}
(x+1)(x+1) & =x^2+x+x+1 \\
& =x^2+2 x+1
\end{aligned}\)
The factorization of the polynomial into a product of two binomials is: x2 + 2x + 1 = (x+1)(x+1).
Page 20 Exercise 9 Answer
Given that the second-order degree polynomial is, x2 + 3x + 2
Here we have to factor the given polynomial into a product of two binomials.
Factorization rule :
Group the first two terms together and then the last two terms together.
Factor out a GCF from each separate binomial.
Factor out the common binomial.
Note that if we multiply our answer out, we do get the original polynomial.
Given Expression: x2+3x+2
factoring out, we get
x2+3x+2=x2+x+2x+2
=x(x+1)+2(x+1)
=(x+2)(x+1)
Check:
Now consider the binomial
⇒ \(\begin{aligned}
(x+2)(x+1) & =x^2+x+2 x+2 \\
& =x^2+3 x+2
\end{aligned}\)
The factorization of the polynomial into a product of two binomials is: x2 + 3x + 2 = (x+1)(x+2).
Page 20 Exercise 10 Answer
Given that the second-order degree polynomial is x2 + 13x + 36
Here we have to factor the given polynomial into a product of two binomials.
Factorization rule :
Group the first two terms together and then the last two terms together.
Factor out a GCF from each separate binomial.
Factor out the common binomial.
Note that if we multiply our answer out, we do get the original polynomial.
Given expression: x2+13x+36
factoring out, we get
⇒ \(\begin{aligned}
x^2+13 x+3 & =x^2+4 x+9 x+36 \\
& =x(x+4)+9(x+4) \\
& =(x+9)(x+4)
\end{aligned}\)
check:
Now consider the binomial
(x+9)(x+4)=x2+4x+9x+36
=x2+13x+36
The factorization of the polynomial into a product of two binomials is x2 + 13x + 36 = (x+4)(x+9).
Page 20 Exercise 12 Answer
Given that the second-order degree polynomial is x2 + 12x + 36
Here we have to factor the given polynomial into a product of two binomials.
Factorization rule :
Group the first two terms together and then the last two terms together.
Factor out a GCF from each separate binomial.
Factor out the common binomial.
Note that if we multiply our answer out, we do get the original polynomial.
Given polynomial: x2+12x+36
factoring out, we get
x^2+12 x+36 & =x^2+6 x+6 x+36 \\
& =x(x+6)+6(x+6) \\
& =(x+6)(x+6)
\end{aligned}\)
or
=(x+6)²
Check:
Now consider the binomial
(x+6)(x+6)= x²+6x+6x+36
=x²+12x+36
The factorization of the polynomial into a product of two binomials is x2 + 12x + 36 = (x+6)(x+6).
Page 20 Exercise 13 Answer
Given that the second-order degree polynomial is x2 + 37x + 36
Here we have to factor the given polynomial into a product of two binomials.
Factorization rule :
Group the first two terms together and then the last two terms together.
Factor out a GCF from each separate binomial.
Factor out the common binomial.
Note that if we multiply our answer out, we do get the original polynomial.
Given polynomial: x²+37x+36
Factoring out, we get
x2+37+36= x2+x+36x+36
=x(x+1)+36(X+1)
=(x+36)(X+1)
Check:
Now consider the binomial,
(x+36)(x+1)=x2+x+36x+36
=x2+37x+36
The factorization of the polynomial into a product of two binomials is x2 + 37x + 36 = (x+11)(x+36).
Page 20 Exercise 14 Answer
Given that the second-order degree polynomial is x2 + 15x + 36
Here we have to factor the given polynomial into a product of two binomials.
Factorization rule :
Group the first two terms together and then the last two terms together.
Factor out a GCF from each separate binomial.
Factor out the common binomial.
Note that if we multiply our answer out, we do get the original polynomial.
Given polynomial: x²+15x+36
Factoring out, we get
⇒ \(\begin{aligned}
x^2+15 x+36 & =x^2+3 x+12 x+36 \\
& =x(x+3)+12(x+3) \\
& =(x+12)(x+3)
\end{aligned}\)
Check:
Now consider the binomial,
\(\begin{aligned}(x+12)(x+3) & =x^2+3 x+12 x+36 \\
& =x^2+15 x+36
\end{aligned}\)
The factorization of the polynomial into a product of two binomials is: x2 + 15x + 36 = (x+3)(x+12).
Page 20 Exercise 15 Answer
Given that the second-order degree polynomial is x2 + 17x + 30
Here we have to factor the given polynomial into a product of two binomials.
Factorization rule :
Group the first two terms together and then the last two terms together.
Factor out a GCF from each separate binomial.
Factor out the common binomial.
Note that if we multiply our answer out, we do get the original polynomial.
Givenpolynomial: x2+17x+30
Factoring out, we get
\(\begin{aligned}x^2+17 x+30 & =x^2+2 x+15 x+30 \\
& =x(x+2)+15(x+2) \\
& =(x+15)(x+2)
\end{aligned}\)
Check:
Now consider the binomials
\(\begin{aligned}(x+15)(x+2) & =x^2+2 x+15 x+30 \\
& =x^2+17 x+30
\end{aligned}\)
The factorization of the polynomial into a product of two binomials is: x2 + 17x + 30 = (x+2)(x+15).
Page 20 Exercise 16 Answer
Given that the second-order degree polynomial is x2 + 11x + 30
Here we have to factor the given polynomial into a product of two binomials.
Factorization rule :
Group the first two terms together and then the last two terms together.
Factor out a GCF from each separate binomial.
Factor out the common binomial.
Note that if we multiply our answer out, we do get the original polynomial.
Given Polynomial: x²+11x+30
Factoring out, we get
\(\begin{aligned}x^2+11 x+30 & =x^2+5 x+6 x+30 \\
& =x(x+5)+6(x+5) \\
& =f(x+6)(x+5)
\end{aligned}\)
Check:
Now consider the binomial:
(x+6)(x+5)=x2+5x+6x+30
=x2+11x+30
The factorization of the polynomial into a product of two binomials is x2 + 11x + 30 = (x+5)(x+6).
Page 20 Exercise 17 Answer
Given that the second-order degree polynomial is x2 + 13x + 30
Here we have to factor the given polynomial into a product of two binomials.
Factorization rule:
Group the first two terms together and then the last two terms together.
Factor out a GCF from each separate binomial.
Factor out the common binomial.
Note that if we multiply our answer out, we do get the original polynomial.
Given polynomial: x2+13x+30
Factoring out, we get
x^2+13 x+30 & =x^2+10 x+3 x+30 \\
& =x(x+10)+3(x+10) \\
& =(x+3)(x+10)
\end{aligned}\)
Check:
Now consider the binomial:
(x+10)(x+3)= x²+3x+10x+30
=x²+13x+30
The factorization of the polynomial into a product of two binomials is: x2 + 13x + 30 = (x+10)(x+3).
Page 20 Exercise 18 Answer
Given: x2 + 31x + 30
To find: factorize each into a product of two binomials.
Given that x²+31x+30
First factorizing as 30 many different ways
we can 30×1, 15×2, 10×3
While testing the combinations, 30×1
So,
x²+31x+30
⇒ x²+x+30x+30
⇒x(x+1)+30(x+1)
⇒(x+1)(x+30)
finally, x²+31x+30=(x+1)(x+30)
Thus, the factorization of the second degree polynomial x2 + 31x + 30 into a product of two binomials is (x+1)(x+30)