In this lesson, we will learn how to divide polynomials by polynomials using factorization and how to find common factors that will cancel out.
Q1:
Find the value of π that makes the expression π₯ β π π₯ + 3 0 2 divisible by π₯ β 5 .
Q2:
Find the value of π that makes the expression π₯ β π π₯ + 2 0 2 divisible by π₯ β 5 .
Q3:
Find the value of π that makes the expression π₯ β π π₯ β 1 8 2 divisible by π₯ + 2 .
Q4:
What is the width of a rectangle whose area is οΉ 4 π₯ β 3 2 π₯ + 6 π₯ ο 3 2 4 cm2 and whose length is οΉ 8 π₯ + 3 π₯ ο 2 cm?
Q5:
What is the width of a rectangle whose area is οΉ 1 4 π₯ + 7 2 π₯ β 8 π₯ ο 4 2 6 cm2 and whose length is οΉ 8 π₯ β 2 π₯ ο 3 cm?
Q6:
What is the width of a rectangle whose area is οΉ 5 9 π₯ β 3 6 π₯ + 7 π₯ ο 5 4 6 cm2 and whose length is οΉ 9 π₯ + π₯ ο 2 3 cm?
Q7:
The area of a triangle is οΉ 1 2 π₯ + 3 8 π₯ + 2 8 ο 2 cm2 and its base is ( 2 π₯ + 4 ) cm. Write an expression for its height.
Q8:
The area of a triangle is οΉ 1 8 π₯ + 7 3 π₯ + 3 5 ο 2 cm2 and its base is ( 2 π₯ + 7 ) cm. Write an expression for its height.
Q9:
The area of a triangle is οΉ 4 8 π₯ + 6 2 π₯ + 9 ο 2 cm2 and its base is ( 8 π₯ + 9 ) cm. Write an expression for its height.
Q10:
A rectangle has an area of οΉ π¦ + 2 π¦ + 5 π¦ + 1 0 ο 3 2 cm2 and a width of ( π¦ + 2 ) cm. Find its length in terms of π¦ and its perimeter when π¦ = 4 .
Q11:
A rectangle has an area of οΉ π¦ + 1 0 π¦ + 7 π¦ + 7 0 ο 3 2 cm2 and a width of ( π¦ + 1 0 ) cm. Find its length in terms of π¦ and its perimeter when π¦ = 2 .
Q12:
Knowing that the length of a rectangle is π₯ + 5 and its area is 2 π₯ + 9 π₯ β 5 2 , express the width of the rectangle algebraically.
Q13:
Knowing that the length of a rectangle is 2 π₯ + 5 and its area is 4 π₯ + 1 0 π₯ + 6 π₯ + 1 5 3 2 , express the width of the rectangle algebraically.
Q14:
Knowing that the volume of a box is 1 0 π₯ + 3 0 π₯ β 8 π₯ β 2 4 3 2 , its length is 2, and its width is π₯ + 3 , express the height of the box algebraically.
Q15:
By factoring, find all the solutions to π₯ β π₯ β 1 4 π₯ + 2 4 = 0 3 2 , given that ( π₯ + 4 ) is a factor of π₯ β π₯ β 1 4 π₯ + 2 4 3 2 .
Q16:
β― + 5 0 π π + 3 8 π π β― = β 2 1 π π + 2 5 π + 1 9 2 .
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