Solve x²-8x+15=3(x-3): Trinomial Decomposition Method

Quadratic Equations with Trinomial Factoring

Solve the equation using trinomial decomposition:

x28x+15=3(x3) x^2-8x+15=3\cdot(x-3)

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Step-by-step video solution

Watch the teacher solve the problem with clear explanations
00:00 Solve
00:03 Now let's factorize using trinomial
00:06 Let's identify the values suitable for B,C
00:10 In the trinomial, we need to find 2 values whose sum equals B
00:16 and their product equals C
00:24 These are the appropriate numbers
00:29 Now let's put these numbers in the trinomial
00:37 Let's arrange the equation so that one side equals 0
01:01 Group by common factor
01:13 According to the factorization, we'll see when each factor in multiplication equals 0
01:18 Isolate the unknown
01:23 This is one solution
01:26 Let's use the same method for the second factor
01:33 Isolate the unknown
01:37 This is the second solution, and both are the answer to the question

Step-by-step written solution

Follow each step carefully to understand the complete solution
1

Understand the problem

Solve the equation using trinomial decomposition:

x28x+15=3(x3) x^2-8x+15=3\cdot(x-3)

2

Step-by-step solution

Let's solve the given equation:

x28x+15=3(x3) x^2-8x+15=3(x-3)

First, let's organize the equation by opening the parentheses (using the extended distribution law) and combining like terms:

x28x+15=3(x3)x28x+15=3x9x28x+153x+9=0x211x+24=0 x^2-8x+15=3(x-3) \\ x^2-8x+15=3x-9 \\ x^2-8x+15-3x+9=0 \\ x^2-11x+24=0

Now we notice that in the resulting equation the coefficient of the squared term is 1, therefore, we can (try to) factor the expression on the left side using quick trinomial factoring:

Let's look for a pair of numbers whose product equals the free term in the expression, and whose sum equals the coefficient of the first-degree term, meaning two numbers m,n m,\hspace{2pt}n that satisfy:

mn=24m+n=11 m\cdot n=24\\ m+n=-11\\ From the first requirement mentioned, that is - from the multiplication, we notice that the product of the numbers we're looking for needs to yield a positive result, therefore we can conclude that both numbers must have the same signs, according to multiplication rules, and now we'll remember that the possible factors of 24 are 6 and 4, 12 and 2, 8 and 3, or 24 and 1. Meeting the second requirement mentioned, along with the fact that the signs of the numbers we're looking for are equal to each other will lead to the conclusion that the only possibility for the two numbers we're looking for is:

{m=8n=3 \begin{cases} m=-8\\ n=-3 \end{cases}

Therefore we can factor the expression on the left side of the equation to:

x211x+24=0(x8)(x3)=0 x^2-11x+24=0 \\ \downarrow\\ (x-8)(x-3)=0

From here we'll remember that the product of expressions will yield 0 only if at least one of the multiplied expressions equals zero,

Therefore we obtain two simple equations and we'll proceed to solve them by isolating the variable in each of them:

x8=0x=8 x-8=0\\ \boxed{x=8}

or:

x3=0x=3 x-3=0\\ \boxed{x=3}

Let's summarize the solution of the equation:

x28x+15=3(x3)x28x+15=3x9x211x+24=0(x8)(x3)=0x8=0x=8x3=0x=3x=8,3 x^2-8x+15=3(x-3) \\ x^2-8x+15=3x-9 \\ x^2-11x+24=0 \\ \downarrow\\ (x-8)(x-3)=0 \\ \downarrow\\ x-8=0\rightarrow\boxed{x=8}\\ x-3=0\rightarrow\boxed{x=3}\\ \downarrow\\ \boxed{x=8,3}

Therefore the correct answer is answer B.

3

Final Answer

8,3 8,3

Key Points to Remember

Essential concepts to master this topic
  • Standard Form: Move all terms to one side to get ax2+bx+c=0 ax^2 + bx + c = 0
  • Factoring: Find two numbers that multiply to 24 and add to -11: (-8) × (-3) = 24
  • Check: Substitute x = 8 and x = 3 back into original equation ✓

Common Mistakes

Avoid these frequent errors
  • Not expanding the right side completely
    Don't leave 3(x-3) unexpanded = incomplete setup! Students often try to factor before getting standard form. Always expand 3(x-3) = 3x - 9 first, then move all terms to one side.

Practice Quiz

Test your knowledge with interactive questions

\( x^2-3x-18=0 \)

FAQ

Everything you need to know about this question

Why do I need to expand 3(x-3) first?

+

You need to see all terms clearly before moving them to one side. Expanding 3(x-3) = 3x - 9 shows you exactly what to subtract from both sides to reach standard form.

How do I find the two numbers for factoring?

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Look for two numbers that multiply to give the constant term (24) and add to give the coefficient of x (-11). Try factor pairs of 24: 1×24, 2×12, 3×8, 4×6.

What if I can't find two numbers that work?

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If no integer pairs work, the trinomial doesn't factor nicely. You'd need to use the quadratic formula instead. But in this case, -8 and -3 work perfectly!

Why are both numbers negative?

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Since their product is positive (24) and their sum is negative (-11), both numbers must have the same sign. The negative sum tells us they're both negative.

How do I check my answers?

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Substitute each solution back into the original equation. For x = 8: 828(8)+15=15 8^2 - 8(8) + 15 = 15 and 3(83)=15 3(8-3) = 15

Can a quadratic equation have two solutions?

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Yes! Most quadratic equations have exactly two solutions. When you factor into (x8)(x3)=0 (x-8)(x-3) = 0 , each factor gives you one solution.

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