Each solution shows how the factors were found and, where useful, a check by multiplying. Compare your method as well as your answer, because many factoring errors are sign mistakes in the middle step.
Problem 1: GCF of a binomial
The GCF of 6 and 9 is 3, and both terms contain \(x\), so the GCF is \(3x\).
\(6x^{2}: 3x=2x\) and \(9x: 3x=3\).
\(3 x (2 x + 3)\)
Problem 2: GCF with powers
The GCF of 12 and 20 is 4. The lowest power of \(a\) is \(a^{2}\). So the GCF is \(4a^{2}\).
\(4 a^{2} (3 a – 5)\)
Problem 3: GCF of a trinomial
The GCF of 8, 12 and 16 is 4, and the lowest power of \(x\) is \(x\), so the GCF is \(4x\).
\(4 x (2 x^{2} + 3 x – 4)\)
Problem 4: Factor x² + bx + c
Find two numbers with product 15 and sum 8: the pairs are 1 and 15 (sum 16) and 3 and 5 (sum 8). Use 3 and 5.
\((x + 3) (x + 5)\)
Problem 5: Both numbers negative
The product 20 is positive and the sum \(-9\) is negative, so both numbers are negative. The pair \(-4\) and \(-5\) has product 20 and sum \(-9\).
\((x – 4) (x – 5)\)
Problem 6: Opposite signs
The product \(-35\) is negative, so the numbers have opposite signs. The pair 7 and \(-5\) has product \(-35\) and sum 2.
\((x + 7) (x – 5)\)
Problem 7: Larger negative number
Find two numbers with product \(-28\) and sum \(-3\). The pair \(-7\) and 4 works: \((-7)(4)=-28\) and \(-7+4=-3\).
\((x – 7) (x + 4)\)
Problem 8: Difference of squares
\(x^{2}-49=x^{2}-7^{2}\), which fits \(a^{2}-b^{2}=(a+b)(a-b)\).
\((x – 7) (x + 7)\)
Problem 9: Difference of squares with a coefficient
\(25x^{2}=(5x)^{2}\) and \(16=4^{2}\), so \(a=5x\) and \(b=4\).
\((5 x – 4) (5 x + 4)\)
Problem 10: Perfect square trinomial
The first term is \(x^{2}\) and the last is \(6^{2}\). The middle term is \(2\cdot x\cdot 6=12x\), so the pattern \(a^{2}+2ab+b^{2}\) applies.
\((x + 6)^{2}\)
Problem 11: Perfect square with a coefficient
\(4x^{2}=(2x)^{2}\) and \(25=5^{2}\). The middle term is \(-2(2x)(5)=-20x\), so the pattern \(a^{2}-2ab+b^{2}\) applies.
\((2 x – 5)^{2}\)
Problem 12: Factor ax² + bx + c
\(ac=2\cdot 3=6\). Two numbers with product 6 and sum 7 are 1 and 6.
Split the middle term: \(2x^{2}+x+6x+3\). Group: \(x(2x+1)+3(2x+1)\).
\((2 x + 1) (x + 3)\)
Problem 13: ax² + bx + c, negative middle
\(ac=24\). Two numbers with product 24 and sum \(-10\) are \(-4\) and \(-6\).
\(3x^{2}-4x-6x+8=x(3x-4)-2(3x-4)\).
\((3 x – 4) (x – 2)\)
Problem 14: ax² + bx + c, negative constant
\(ac=-90\). Two numbers with product \(-90\) and sum 1 are 10 and \(-9\).
\(6x^{2}+10x-9x-15=2x(3x+5)-3(3x+5)\).
\((2 x – 3) (3 x + 5)\)
Problem 15: Factor by grouping
Group the first two and last two terms: \((x^{3}+4x^{2})+(3x+12)\). Factor each group: \(x^{2}(x+4)+3(x+4)\).
\((x^{2} + 3) (x + 4)\)
Problem 16: GCF then difference of squares
Pull out the GCF 2: \(2(x^{2}-16)\). Then \(x^{2}-16=(x-4)(x+4)\).
\(2 (x – 4) (x + 4)\)
Problem 17: GCF then perfect square
The GCF is \(3x\): \(3x(x^{2}+6x+9)\). The trinomial is a perfect square: \(x^{2}+6x+9=(x+3)^{2}\).
\(3 x (x + 3)^{2}\)
Problem 18: Zeros of a quadratic
Find two numbers with product \(-14\) and sum \(-5\): \(-7\) and 2.
\((x – 7) (x + 2)\) \(=0\). By the zero product property, \(x-7=0\) or \(x+2=0\).
\(x=7\) or \(x=-2\)
Problem 19: Zeros with a leading coefficient
\(ac=-12\). Two numbers with product \(-12\) and sum 1 are 4 and \(-3\).
\(2x^{2}+4x-3x-6=2x(x+2)-3(x+2)=(2x-3)(x+2)\).
Set each factor to zero: \(2x-3=0\) gives \(x=\frac{3}{2}\), and \(x+2=0\) gives \(x=-2\).
\(x=\frac{3}{2}\) or \(x=-2\)
Problem 20: Dimensions of a rectangle
Numbers with product 14 and sum 9 are 2 and 7, so the area factors as \((x + 2) (x + 7)\). The width is \(x+2\) and the length is \(x+7\).
If \(x+2=5\), then \(x=3\), so the length is \(3+7=10\) feet and the area is \(5\times 10=50\) square feet. Check: \(9+27+14=50\).
length 10 ft, area 50 square feet
Problem 21: When does the ball land?
Pull out the GCF \(-16\): \(-16(t^{2}-4t-5)\). Then \(t^{2}-4t-5=(t-5)(t+1)\).
\(-16(t-5)(t+1)=0\) gives \(t=5\) or \(t=-1\). Time cannot be negative, so the ball lands after 5 seconds.
Check: \(-16(25)+64(5)+80=-400+320+80=0\).
Problem 22: Consecutive integers
Let the integers be \(n\) and \(n+1\). Then \(n(n+1)=156\), which gives \(n^{2}+n-156=0\).
Numbers with product \(-156\) and sum 1: 13 and \(-12\). So \((n+13)(n-12)=0\), and \(n=-13\) or \(n=12\).
The integers are positive, so \(n=12\). Check: \(12\times 13=156\).
12 and 13
Problem 23: Is it factored completely?
The product is correct: \((2x)^{2}-6^{2}=4x^{2}-36\). But each binomial still has a common factor of 2: \(2x-6=2(x-3)\) and \(2x+6=2(x+3)\).
A better first step is the GCF: \(4(x^{2}-9)\).
\(4 (x – 3) (x + 3)\)
Problem 24: Banner dimensions
\(ac=36\). Two numbers with product 36 and sum 13 are 4 and 9.
\(6x^{2}+4x+9x+6=2x(3x+2)+3(3x+2)\).
\((2 x + 3) (3 x + 2)\)
See every chapter on the Algebra 1 math page.