Problem 1 :
x2 + 2x + xy + 2y
Problem 2 :
2ax2 + bx2 - 2ay2 - by2
Problem 3 :
x3 + x2 - x - 1
Problem 4 :
2x2 - 8
Problem 5 :
2x - 2xy2
Problem 6 :
3t3 - 27t
Problem 7 :
4x4 - 4x2
Problem 8 :
3x + x2 - 10
Problem 9 :
x3 + 8x2 - x - 8
Problem 10 :
4a3 - 49a
Problem 11 :
2y2 - 16y + 32
Problem 12 :
p2 q - 25q + 3q2 - 75
Problem 13 :
16 - w4
Problem 14 :
64x3 + 27
Problem 15 :
3t3 - 27t
Problem 1 :
x2 + 2x + xy + 2y
Solution :
= x2 + 2x + xy + 2y
Since we have four terms in the given expression, we have to use the method of grouping and do the factoring.
Factoring x from first two terms and factoring y from third and fourth terms.
= x (x + 2) + y (x + 2)
= (x + y)(x + 2)
So, the factored form of the given expression is
(x + y)(x + 2)
Problem 2 :
2ax2 + bx2 - 2ay2 - by2
Solution :
= 2ax2 + bx2 - 2ay2 - by2
Since we have four terms in the given expression, we have to use grouping method.
Factoring x2 from first and second term, factoring y2 from third and fourth term.
= x2 (2a + b) - y2 (2a + b)
= (x2 - y2) (2a + b)
So, the factored form of the given expressions is
(x2 - y2) (2a + b)
Problem 3 :
x3 + x2 - x - 1
Solution :
= x3 + x2 - x - 1
Since we have four terms in the given expression, we have to use grouping method.
Factoring x2 from first and second term, factoring -1 from third and fourth terms.
= x2 (x + 1) - 1(x + 1)
= (x2 - 1)(x + 1)
Here (x2 - 1) looks like a2 - b2, by expanding this using algebraic identity, we get (a + b) (a - b)
= (x + 1)(x - 1)(x + 1)
Since x + 1 is repeating twice, we write in the exponential form
= (x + 1)2(x - 1)
Problem 4 :
2x2 - 8
Solution :
= 2x2 - 8
= 2(x2 - 4)
= 2(x2 - 22)
= 2(x + 2)(x - 2)
So, the factored form of the expression is 2(x + 2)(x - 2).
Problem 5 :
2x - 2xy2
Solution :
= 2x - 2xy2
Factoring 2x, we get
= 2x (1 - y2)
Here 1 - y2, looks like a2 - b2 then (a + b) (a - b)
= 2x(1 + y)(1 - y)
Problem 6 :
3t3 - 27t
Solution :
= 3t3 - 27t
Factoring 3t, we get
= 3t(t2 - 9)
= 3t(t2 - 32)
Here t2 - 32, looks like a2 - b2 then (a + b) (a - b)
= 3t (t + 3)(t - 3)
So, the factored form of the given expression is
3t (t + 3)(t - 3)
Problem 7 :
4x4 - 4x2
Solution :
Factoring 4x2 from these two terms
= 4x2 (x2 - 1)
Here x2 - 12, looks like a2 - b2 then (x + 1)(x - 1)
= 4x2 (x + 1)(x - 1)
Problem 8 :
3x + x2 - 10
Solution :
3x + x2 - 10
It is a trinomial, but it is not in the standard form.
= x2 + 3x - 10
By splitting the middle term, we get
= x2 + 5x - 2x - 10
= x(x + 5) - 2(x + 5)
= (x - 2)(x + 5)
So, the factored form of the given expression is
(x - 2)(x + 5)
Problem 9 :
x3 + 8x2 - x - 8
Solution :
= x3 + 8x2 - x - 8
= x2(x + 8) - 1(x + 8)
= (x2 - 1)(x + 8)
= (x + 1)(x - 1)(x + 8)
So, the factored form of the given expression is
(x + 1)(x - 1)(x + 8)
Problem 10 :
4a3 - 49a
Solution :
= 4a3 - 49a
Factoring a from the given expression.
= a(4a2 - 49)
= a((2a)2 - 72)
= a(2a + 7)(2a - 7)
So, the factored form of the given expression is
a(2a + 7)(2a - 7)
Problem 11 :
2y2 - 16y + 32
Solution :
= 2y2 - 16y + 32
Factoring 2, we get
= 2(y2 - 8y + 16)
= 2(y2 - 4y - 4y + 16)
= 2[y(y - 4) - 4(y - 4)]
= 2(y - 4)(y - 4)
So, the factored form of the given expression is
2(y - 4)(y - 4)
Problem 12 :
p2 q - 25q + 3q2 - 75
Solution :
= p2 q - 25q + 3q2 - 75
= q(p2 - 25) + 3(p2 - 25)
= (q + 3)(p2 - 25)
Here (p2 - 25) looks like a2 - b2, expanding this using algebraic identity, we get
= (q + 3)(q + 5)(q - 5)
Problem 13 :
16 - w4
Solution :
= 16 - w4
= 42 - (w2)2
= (4 + w2) (4 - w2)
= (4 + w2) (22 - w2)
= (4 + w2) (2 + w)(2 - w)
So, the factored form of the given expression is
(4 + w2) (2 + w)(2 - w)
Problem 14 :
64x3 + 27
Solution :
= 64x3 + 27
= (4x)3 + 33
Looks like a3 + b3 = (a + b) (a2 - ab + b2)
= (4x + 3)((4x)2 - 4x(3) + 32)
= (4x + 3)(16x2 - 12x + 9)
Problem 15 :
3t3 - 27t
Solution :
= 3t3 - 27t
= 3t(t2 - 9)
= 3t(t2 - 32)
= 3t(t + 3)(t - 3)
So, the factored form of the expression is
3t(t + 3)(t - 3)
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