ਸਿੱਖਣ ਦੇ ਉਦੇਸ਼
ਇਸ ਭਾਗ ਦੇ ਅੰਤ ਤੱਕ, ਤੁਸੀਂ ਇਹ ਕਰਨ ਦੇ ਯੋਗ ਹੋਵੋਗੇ:
x^2+bx+c ਰੂਪ ਦੇ ਤ੍ਰਿਪਦੀਆਂ ਦਾ ਅੰਸ਼ਕ ਗੁਣਨਖੰਡਨ ਕਰਨਾ
a x^2+bx+c ਰੂਪ ਦੇ ਤ੍ਰਿਪਦੀਆਂ ਦਾ ਅੰਸ਼ਕ ਗੁਣਨਖੰਡਨ ਕਰਨਾ, ਅਜ਼ਮਾਇਸ਼ ਅਤੇ ਤਰੁੱਟੀ ਵਿਧੀ ਨਾਲ
a x^2+bx+c ਰੂਪ ਦੇ ਤ੍ਰਿਪਦੀਆਂ ਦਾ ਅੰਸ਼ਕ ਗੁਣਨਖੰਡਨ ਕਰਨਾ, 'ac' ਵਿਧੀ ਨਾਲ
ਬਦਲਵੀਂ ਵਰਤੋਂ ਨਾਲ ਅੰਸ਼ਕ ਗੁਣਨਖੰਡਨ ਕਰਨਾ
ਤਿਆਰ ਰਹੋ 6.4
ਸ਼ੁਰੂ ਕਰਨ ਤੋਂ ਪਹਿਲਾਂ, ਇਹ ਤਿਆਰੀ ਕਵਿਜ਼ ਲਓ।
72 ਦੇ ਸਾਰੇ ਗੁਣਨਖੰਡ ਲੱਭੋ। ਜੇਕਰ ਤੁਸੀਂ ਇਹ ਸਮੱਸਿਆ ਗੁਆਚੀ ਹੈ, ਤਾਂ ਉਦਾਹਰਨ 1.2 ਦੀ ਸਮੀਖਿਆ ਕਰੋ।
ਤਿਆਰ ਰਹੋ 6.5
ਗੁਣਨਫਲ ਲੱਭੋ: (3y+4)(2y+5)। ਜੇਕਰ ਤੁਸੀਂ ਇਹ ਸਮੱਸਿਆ ਗੁਆਚੀ ਹੈ, ਤਾਂ ਉਦਾਹਰਨ 5.28 ਦੀ ਸਮੀਖਿਆ ਕਰੋ।
ਤਿਆਰ ਰਹੋ 6.6
ਸਰਲ ਕਰੋ: −9(6); −9(−6)। ਜੇਕਰ ਤੁਸੀਂ ਇਹ ਸਮੱਸਿਆ ਗੁਆਚੀ ਹੈ, ਤਾਂ ਉਦਾਹਰਨ 1.18 ਦੀ ਸਮੀਖਿਆ ਕਰੋ।
x^2+bx+c ਰੂਪ ਦੇ ਤ੍ਰਿਪਦੀਆਂ ਦਾ ਅੰਸ਼ਕ ਗੁਣਨਖੰਡਨ
ਤੁਸੀਂ ਪਹਿਲਾਂ ਹੀ FOIL ਦੀ ਵਰਤੋਂ ਕਰਕੇ ਦੋਪਦੀਆਂ ਨੂੰ ਗੁਣਾ ਕਰਨਾ ਸਿੱਖ ਲਿਆ ਹੈ। ਹੁਣ ਤੁਹਾਨੂੰ ਇਸ ਗੁਣਾ ਨੂੰ "ਉਲਟਾ" ਕਰਨਾ ਪਵੇਗਾ। ਤ੍ਰਿਪਦੀ ਦਾ ਅੰਸ਼ਕ ਗੁਣਨਖੰਡਨ ਕਰਨ ਦਾ ਮਤਲਬ ਹੈ ਗੁਣਨਫਲ ਤੋਂ ਸ਼ੁਰੂ ਕਰਨਾ, ਅਤੇ ਗੁਣਨਖੰਡਾਂ ਨਾਲ ਖਤਮ ਕਰਨਾ।
ਇਹ ਪਤਾ ਲਗਾਉਣ ਲਈ ਕਿ ਅਸੀਂ x^2+bx+c, ਜਿਵੇਂ ਕਿ x^2+5x+6, ਰੂਪ ਦੀ ਤ੍ਰਿਪਦੀ ਦਾ ਅੰਸ਼ਕ ਗੁਣਨਖੰਡਨ ਕਿਵੇਂ ਕਰਾਂਗੇ ਅਤੇ ਇਸਨੂੰ (x+2)(x+3) ਵਿੱਚ ਕਿਵੇਂ ਵੰਡਾਂਗੇ, ਆਓ (x+m) ਅਤੇ (x+n) ਰੂਪ ਦੀਆਂ ਦੋ ਆਮ ਦੋਪਦੀਆਂ ਤੋਂ ਸ਼ੁਰੂ ਕਰੀਏ।
row: ਗੁਣਨਫਲ ਲੱਭਣ ਲਈ Foil।
row: ਵਿਚਕਾਰਲੇ ਪਦਾਂ ਤੋਂ GCF ਦਾ ਅੰਸ਼ਕ ਗੁਣਨਖੰਡਨ ਕਰੋ।
row: ਸਾਡੀ ਤ੍ਰਿਪਦੀ x^2+bx+c ਰੂਪ ਦੀ ਹੈ।
ਇਹ ਸਾਨੂੰ ਦੱਸਦਾ ਹੈ ਕਿ x^2+bx+c ਰੂਪ ਦੀ ਤ੍ਰਿਪਦੀ ਦਾ ਅੰਸ਼ਕ ਗੁਣਨਖੰਡਨ ਕਰਨ ਲਈ, ਸਾਨੂੰ ਦੋ ਗੁਣਨਖੰਡ (x+m) ਅਤੇ (x+n) ਦੀ ਲੋੜ ਹੈ ਜਿੱਥੇ ਦੋ ਸੰਖਿਆਵਾਂ m ਅਤੇ n ਦਾ ਗੁਣਨਫਲ c ਹੋਵੇ ਅਤੇ ਜੋੜ b ਹੋਵੇ।
ਉਦਾਹਰਨ 6.9
x^2+bx+c ਰੂਪ ਦੀ ਤ੍ਰਿਪਦੀ ਦਾ ਅੰਸ਼ਕ ਗੁਣਨਖੰਡਨ ਕਿਵੇਂ ਕਰਨਾ ਹੈ
ਅੰਸ਼ਕ ਗੁਣਨਖੰਡਨ ਕਰੋ: x^2+11x+24।
ਹੱਲ
Try It 6.17
Factor: q2+10q+24.q2+10q+24.
Try It 6.18
Factor: t2+14t+24.t2+14t+24.
Let’s summarize the steps we used to find the factors.
How To
Factor trinomials of the form x2+bx+c.x2+bx+c.
Step 1. Write the factors as two binomials with first terms x. x2+bx+c(x)(x)x2+bx+c(x)(x)
Step 2. Find two numbers m and n that multiply to c,m·n=cc,m·n=c add to b,m+n=bb,m+n=b
Step 3. Use m and n as the last terms of the factors. (x+m)(x+n)(x+m)(x+n)
Step 4. Check by multiplying the factors.
In the first example, all terms in the trinomial were positive. What happens when there are negative terms? Well, it depends which term is negative. Let’s look first at trinomials with only the middle term negative.
How do you get a positive product and a negative sum? We use two negative numbers.
Example 6.10
Factor: y2−11y+28.y2−11y+28.
Solution
Again, with the positive last term, 28, and the negative middle term, −11y,−11y, we need two negative factors. Find two numbers that multiply 28 and add to −11.−11.
row: y2−11y+28y2−11y+28
row: Write the factors as two binomials with first terms y.y. | (y)(y)(y)(y)
row: Find two numbers that: multiply to 28 and add to −11.
row: Factors of 2828 | Sum of factors
row: −1,−28−1,−28−2,−14−2,−14−4,−7−4,−7 | −1+(−28)=−29−1+(−28)=−29−2+(−14)=−16−2+(−14)=−16−4+(−7)=−11*−4+(−7)=−11*
row: Use −4,−7−4,−7 as the last terms of the binomials. | (y−4)(y−7)(y−4)(y−7)
row: Check: (y−4)(y−7)y2−7y−4y+28y2−11y+28✓(y−4)(y−7)y2−7y−4y+28y2−11y+28✓
Try It 6.19
Factor: u2−9u+18.u2−9u+18.
Try It 6.20
Factor: y2−16y+63.y2−16y+63.
Now, what if the last term in the trinomial is negative? Think about FOIL. The last term is the product of the last terms in the two binomials. A negative product results from multiplying two numbers with opposite signs. You have to be very careful to choose factors to make sure you get the correct sign for the middle term, too.
How do you get a negative product and a positive sum? We use one positive and one negative number.
When we factor trinomials, we must have the terms written in descending order—in order from highest degree to lowest degree.
Example 6.11
Factor: 2x+x2−48.2x+x2−48.
Solution
row: 2x+x2−482x+x2−48
row: First we put the terms in decreasing degree order. | x2+2x−48x2+2x−48
row: Factors will be two binomials with first terms x.x. | (x)(x)(x)(x)
row: Factors of −48−48 | Sum of factors
row: −1,48−1,48−2,24−2,24−3,16−3,16−4,12−4,12−6,8−6,8 | −1+48=47−1+48=47−2+24=22−2+24=22−3+16=13−3+16=13−4+12=8−4+12=8−6+8=2*−6+8=2*
row: Use−6,8as the last terms of the binomials.Use−6,8as the last terms of the binomials. | (x−6)(x+8)(x−6)(x+8)
row: Check: (x−6)(x+8)x2−6q+8q−48x2+2x−48✓(x−6)(x+8)x2−6q+8q−48x2+2x−48✓
Try It 6.21
Factor: 9m+m2+18.9m+m2+18.
Try It 6.22
Factor: −7n+12+n2.−7n+12+n2.
Sometimes you’ll need to factor trinomials of the form x2+bxy+cy2x2+bxy+cy2 with two variables, such as x2+12xy+36y2.x2+12xy+36y2. The first term, x2,x2, is the product of the first terms of the binomial factors, x·x.x·x. The y2y2 in the last term means that the second terms of the binomial factors must each contain y. To get the coefficients b and c, you use the same process summarized in How To Factor trinomials.
Example 6.12
Factor: r2−8rs−9s2.r2−8rs−9s2.
Solution
We need r in the first term of each binomial and s in the second term. The last term of the trinomial is negative, so the factors must have opposite signs.
row: r2−8rs−9s2r2−8rs−9s2
row: Note that the first terms are r,r, last terms contain s.s. | (rs)(rs)(rs)(rs)
row: Find the numbers that multiply to −9 and add to −8.
row: Factors of −9−9 | Sum of factors
row: 1,−91,−9 | −1+9=8−1+9=8
row: −1,9−1,9 | 1+(−9)=−8*1+(−9)=−8*
row: 3,−33,−3 | 3+(−3)=03+(−3)=0
row: Use1,−9as coefficients of the last terms.Use1,−9as coefficients of the last terms. | (r+s)(r−9s)(r+s)(r−9s)
row: Check: (r−9s)(r+s)r2+rs−9rs−9s2r2−8rs−9s2✓(r−9s)(r+s)r2+rs−9rs−9s2r2−8rs−9s2✓
Try It 6.23
Factor: a2−11ab+10b2.a2−11ab+10b2.
Try It 6.24
Factor: m2−13mn+12n2.m2−13mn+12n2.
Some trinomials are prime. The only way to be certain a trinomial is prime is to list all the possibilities and show that none of them work.
Example 6.13
Factor: u2−9uv−12v2.u2−9uv−12v2.
Solution
We need u in the first term of each binomial and v in the second term. The last term of the trinomial is negative, so the factors must have opposite signs.
row: u2−9uv−12v2u2−9uv−12v2
row: Note that the first terms are u,u, last terms contain v.v. | (uv)(uv)(uv)(uv)
row: Find the numbers that multiply to −12 and add to −9.
row: Factors of −12−12 | Sum of factors
row: 1,−121,−12−1,12−1,122,−62,−6−2,6−2,63,−43,−4−3,4−3,4 | 1+(−12)=−111+(−12)=−11−1+12=11−1+12=112+(−6)=−42+(−6)=−4−2+6=4−2+6=43+(−4)=−13+(−4)=−1−3+4=1−3+4=1
Note there are no factor pairs that give us −9−9 as a sum. The trinomial is prime.
Try It 6.25
Factor: x2−7xy−10y2.x2−7xy−10y2.
Try It 6.26
Factor: p2+15pq+20q2.p2+15pq+20q2.
Let’s summarize the method we just developed to factor trinomials of the form x2+bx+c.x2+bx+c.
Strategy for Factoring Trinomials of the Form x 2 + b x + c x 2 + b x + c
When we factor a trinomial, we look at the signs of its terms first to determine the signs of the binomial factors.
Notice that, in the case when m and n have opposite signs, the sign of the one with the larger absolute value matches the sign of b.
Factor Trinomials of the form ax2 + bx + c using Trial and Error
Our next step is to factor trinomials whose leading coefficient is not 1, trinomials of the form ax2+bx+c.ax2+bx+c.
Remember to always check for a GCF first! Sometimes, after you factor the GCF, the leading coefficient of the trinomial becomes 1 and you can factor it by the methods we’ve used so far. Let’s do an example to see how this works.
Example 6.14
Factor completely: 4x3+16x2−20x.4x3+16x2−20x.
Solution
row: Is there a greatest common factor? | 4x3+16x2−20x4x3+16x2−20x
row: Yes, GCF=4x.GCF=4x. Factor it. | 4x(x2+4x−5)4x(x2+4x−5)
row: Binomial, trinomial, or more than three terms?
row: It is a trinomial. So “undo FOIL.” | 4x(x)(x)4x(x)(x)
row: Use a table like the one shown to find two numbers thatmultiply to −5 and add to 4. | 4x(x−1)(x+5)4x(x−1)(x+5)
row: Factors of −5−5 | Sum of factors
row: −1,5−1,51,−51,−5 | −1+5=4*−1+5=4*1+(−5)=−41+(−5)=−4
row: Check:4x(x−1)(x+5)4x(x2+5x−x−5)4x(x2+4x−5)4x3+16x2−20x✓4x(x−1)(x+5)4x(x2+5x−x−5)4x(x2+4x−5)4x3+16x2−20x✓
Try It 6.27
Factor completely: 5x3+15x2−20x.5x3+15x2−20x.
Try It 6.28
Factor completely: 6y3+18y2−60y.6y3+18y2−60y.
What happens when the leading coefficient is not 1 and there is no GCF? There are several methods that can be used to factor these trinomials. First we will use the Trial and Error method.
Let’s factor the trinomial 3x2+5x+2.3x2+5x+2.
From our earlier work, we expect this will factor into two binomials.
We know the first terms of the binomial factors will multiply to give us 3x2.3x2. The only factors of 3x23x2 are 1x,3x.1x,3x. We can place them in the binomials.
Check: Does 1x·3x=3x2?1x·3x=3x2?
We know the last terms of the binomials will multiply to 2. Since this trinomial has all positive terms, we only need to consider positive factors. The only factors of 2 are 1, 2. But we now have two cases to consider as it will make a difference if we write 1, 2 or 2, 1.
Which factors are correct? To decide that, we multiply the inner and outer terms.
Since the middle term of the trinomial is 5x,5x, the factors in the first case will work. Let’s use FOIL to check.
Our result of the factoring is:
Example 6.15
How to Factor a Trinomial Using Trial and Error
Factor completely using trial and error: 3y2+22y+7.3y2+22y+7.
Solution
Try It 6.29
Factor completely using trial and error: 2a2+5a+3.2a2+5a+3.
Try It 6.30
Factor completely using trial and error: 4b2+5b+1.4b2+5b+1.
How To
Factor trinomials of the form ax2+bx+cax2+bx+c using trial and error.
Step 1. Write the trinomial in descending order of degrees as needed.
Step 2. Factor any GCF.
Step 3. Find all the factor pairs of the first term.
Step 4. Find all the factor pairs of the third term.
Step 5. Test all the possible combinations of the factors until the correct product is found.
Step 6. Check by multiplying.
Remember, when the middle term is negative and the last term is positive, the signs in the binomials must both be negative.
Example 6.16
Factor completely using trial and error: 6b2−13b+5.6b2−13b+5.
Solution
row: The trinomial is already in descending order.
row: Find the factors of the first term.
row: Find the factors of the last term. Consider the signs.Since the last term, 5, is positive its factors must both bepositive or both be negative. The coefficient of themiddle term is negative, so we use the negative factors.
Consider all the combinations of factors.
row: 6b2−13b+56b2−13b+5
row: Possible factors | Product
row: (b−1)(6b−5)(b−1)(6b−5) | 6b2−11b+56b2−11b+5
row: (b−5)(6b−1)(b−5)(6b−1) | 6b2−31b+56b2−31b+5
row: (2b−1)(3b−5)(2b−1)(3b−5) | 6b2−13b+5*6b2−13b+5*
row: (2b−5)(3b−1)(2b−5)(3b−1) | 6b2−17b+56b2−17b+5
row: The correct factors are those whose productis the original trinomial. | (2b−1)(3b−5)(2b−1)(3b−5)
row: Check by multiplying:(2b−1)(3b−5)6b2−10b−3b+56b2−13b+5✓(2b−1)(3b−5)6b2−10b−3b+56b2−13b+5✓
Try It 6.31
Factor completely using trial and error: 8x2−14x+3.8x2−14x+3.
Try It 6.32
Factor completely using trial and error: 10y2−37y+7.10y2−37y+7.
When we factor an expression, we always look for a greatest common factor first. If the expression does not have a greatest common factor, there cannot be one in its factors either. This may help us eliminate some of the possible factor combinations.
Example 6.17
Factor completely using trial and error: 18x2−37xy+15y2.18x2−37xy+15y2.
Solution
row: The trinomial is already in descending order.
row: Find the factors of the first term.
row: Find the factors of the last term. Consider the signs.Since 15 is positive and the coefficient of the middleterm is negative, we use the negative factors.
Consider all the combinations of factors.
row: The correct factors are those whose product is the original trinomial. | (2x−3y)(9x−5y)(2x−3y)(9x−5y)
row: Check by multiplying: (2x−3y)(9x−5y)18x2−10xy−27xy+15y218x2−37xy+15y2✓(2x−3y)(9x−5y)18x2−10xy−27xy+15y218x2−37xy+15y2✓
Try It 6.33
Factor completely using trial and error 18x2−3xy−10y2.18x2−3xy−10y2.
Try It 6.34
Factor completely using trial and error: 30x2−53xy−21y2.30x2−53xy−21y2.
Don’t forget to look for a GCF first and remember if the leading coefficient is negative, so is the GCF.
Example 6.18
Factor completely using trial and error: −10y4−55y3−60y2.−10y4−55y3−60y2.
Solution
row: Notice the greatest common factor, so factor it first.
row: Factor the trinomial.
Consider all the combinations.
row: The correct factors are those whose productis the original trinomial. Remember to includethe factor −5y2.−5y2. | −5y2(y+4)(2y+3)−5y2(y+4)(2y+3)
row: Check by multiplying: −5y2(y+4)(2y+3)−5y2(2y2+8y+3y+12)−10y4−55y3−60y2✓−5y2(y+4)(2y+3)−5y2(2y2+8y+3y+12)−10y4−55y3−60y2✓
Try It 6.35
Factor completely using trial and error: 15n3−85n2+100n.15n3−85n2+100n.
Try It 6.36
Factor completely using trial and error: 56q3+320q2−96q.56q3+320q2−96q.
Factor Trinomials of the Form ax2+bx+cax2+bx+c using the “ac” Method
Another way to factor trinomials of the form ax2+bx+cax2+bx+c is the “ac” method. (The “ac” method is sometimes called the grouping method.) The “ac” method is actually an extension of the methods you used in the last section to factor trinomials with leading coefficient one. This method is very structured (that is step-by-step), and it always works!
Example 6.19
How to Factor Trinomials using the “ac” Method
Factor using the ‘ac’ method: 6x2+7x+2.6x2+7x+2.
Solution
Try It 6.37
Factor using the ‘ac’ method: 6x2+13x+2.6x2+13x+2.
Try It 6.38
Factor using the ‘ac’ method: 4y2+8y+3.4y2+8y+3.
The “ac” method is summarized here.
How To
Factor trinomials of the form ax2+bx+cax2+bx+c using the “ac” method.
Step 1. Factor any GCF.
Step 2. Find the product ac.
Step 3. Find two numbers m and n that: Multiply toacm·n=a·cAdd tobm+n=bax2+bx+cMultiply toacm·n=a·cAdd tobm+n=bax2+bx+c
Step 4. Split the middle term using m and n. ax2+mx+nx+cax2+mx+nx+c
Step 5. Factor by grouping.
Step 6. Check by multiplying the factors.
Don’t forget to look for a common factor!
Example 6.20
Factor using the ‘ac’ method: 10y2−55y+70.10y2−55y+70.
Solution
row: Is there a greatest common factor?
row: Yes. The GCF is 5.
row: Factor it.
row: The trinomial inside the parentheses has aleading coefficient that is not 1.
row: Find the product ac.ac. | ac=28ac=28
row: Find two numbers that multiply to acac | (−4)(−7)=28(−4)(−7)=28
row: and add to b. | −4+(−7)=−11−4+(−7)=−11
row: Split the middle term.
row: Factor the trinomial by grouping.
row: Check by multiplying all three factors.5(y−2)(2y−7)5(2y2−7y−4y+14)5(2y2−11y+14)10y2−55y+70✓5(y−2)(2y−7)5(2y2−7y−4y+14)5(2y2−11y+14)10y2−55y+70✓
Try It 6.39
Factor using the ‘ac’ method: 16x2−32x+12.16x2−32x+12.
Try It 6.40
Factor using the ‘ac’ method: 18w2−39w+18.18w2−39w+18.
Factor Using Substitution
Sometimes a trinomial does not appear to be in the ax2+bx+cax2+bx+c form. However, we can often make a thoughtful substitution that will allow us to make it fit the ax2+bx+cax2+bx+c form. This is called factoring by substitution. It is standard to use u for the substitution.
In the ax2+bx+c,ax2+bx+c, the middle term has a variable, x, and its square, x2,x2, is the variable part of the first term. Look for this relationship as you try to find a substitution.
Example 6.21
Factor by substitution: x4−4x2−5.x4−4x2−5.
Solution
The variable part of the middle term is x2x2 and its square, x4,x4, is the variable part of the first term. (We know (x2)2=x4).(x2)2=x4). If we let u=x2,u=x2, we can put our trinomial in the ax2+bx+cax2+bx+c form we need to factor it.
row: Rewrite the trinomial to prepare for the substitution.
row: Let u=x2u=x2 and substitute.
row: Factor the trinomial.
row: Replace u with x2.x2.
row: Check:(x2+1)(x2−5)x4−5x2+x2−5x4−4x2−5✓(x2+1)(x2−5)x4−5x2+x2−5x4−4x2−5✓
Try It 6.41
Factor by substitution: h4+4h2−12.h4+4h2−12.
Try It 6.42
Factor by substitution: y4−y2−20.y4−y2−20.
Sometimes the expression to be substituted is not a monomial.
Example 6.22
Factor by substitution: (x−2)2+7(x−2)+12(x−2)2+7(x−2)+12
Solution
The binomial in the middle term, (x−2)(x−2) is squared in the first term. If we let u=x−2u=x−2 and substitute, our trinomial will be in ax2+bx+cax2+bx+c form.
row: Rewrite the trinomial to prepare for the substitution.
row: Let u=x−2u=x−2 and substitute.
row: Factor the trinomial.
row: Replace u with x−2.x−2.
row: Simplify inside the parentheses.
This could also be factored by first multiplying out the (x−2)2(x−2)2 and the 7(x−2)7(x−2) and then combining like terms and then factoring. Most students prefer the substitution method.
Try It 6.43
Factor by substitution: (x−5)2+6(x−5)+8.(x−5)2+6(x−5)+8.
Try It 6.44
Factor by substitution: (y−4)2+8(y−4)+15.(y−4)2+8(y−4)+15.
Media
Access this online resource for additional instruction and practice with factoring.
Factor a trinomial using the AC method
Practice Makes Perfect
Factor Trinomials of the Form x2+bx+cx2+bx+c
In the following exercises, factor each trinomial of the form x2+bx+c.x2+bx+c.
p 2 + 11 p + 30 p 2 + 11 p + 30
w 2 + 10 w + 21 w 2 + 10 w + 21
n 2 + 19 n + 48 n 2 + 19 n + 48
b 2 + 14 b + 48 b 2 + 14 b + 48
a 2 + 25 a + 100 a 2 + 25 a + 100
u 2 + 101 u + 100 u 2 + 101 u + 100
x 2 − 8 x + 12 x 2 − 8 x + 12
q 2 − 13 q + 36 q 2 − 13 q + 36
y 2 − 18 y + 45 y 2 − 18 y + 45
m 2 − 13 m + 30 m 2 − 13 m + 30
x 2 − 8 x + 7 x 2 − 8 x + 7
y 2 − 5 y + 6 y 2 − 5 y + 6
5 p − 6 + p 2 5 p − 6 + p 2
6 n − 7 + n 2 6 n − 7 + n 2
8 − 6 x + x 2 8 − 6 x + x 2
7 x + x 2 + 6 7 x + x 2 + 6
x 2 − 12 − 11 x x 2 − 12 − 11 x
−11 − 10 x + x 2 −11 − 10 x + x 2
In the following exercises, factor each trinomial of the form x2+bxy+cy2.x2+bxy+cy2. If the trinomial cannot be factored, answer “Prime.”
x 2 − 2 x y − 80 y 2 x 2 − 2 x y − 80 y 2
p 2 − 8 p q − 65 q 2 p 2 − 8 p q − 65 q 2
m 2 − 64 m n − 65 n 2 m 2 − 64 m n − 65 n 2
p 2 − 2 p q − 35 q 2 p 2 − 2 p q − 35 q 2
a 2 + 5 a b − 24 b 2 a 2 + 5 a b − 24 b 2
r 2 + 3 r s − 28 s 2 r 2 + 3 r s − 28 s 2
x 2 − 3 x y − 14 y 2 x 2 − 3 x y − 14 y 2
u 2 − 8 u v − 24 v 2 u 2 − 8 u v − 24 v 2
m 2 − 5 m n + 30 n 2 m 2 − 5 m n + 30 n 2
c 2 − 7 c d + 18 d 2 c 2 − 7 c d + 18 d 2
Factor Trinomials of the Form ax2+bx+cax2+bx+c Using Trial and Error
In the following exercises, factor completely using trial and error.
p 3 − 8 p 2 − 20 p p 3 − 8 p 2 − 20 p
q 3 − 5 q 2 − 24 q q 3 − 5 q 2 − 24 q
3 m 3 − 21 m 2 + 30 m 3 m 3 − 21 m 2 + 30 m
11 n 3 − 55 n 2 + 44 n 11 n 3 − 55 n 2 + 44 n
5 x 4 + 10 x 3 − 75 x 2 5 x 4 + 10 x 3 − 75 x 2
6 y 4 + 12 y 3 − 48 y 2 6 y 4 + 12 y 3 − 48 y 2
2 t 2 + 7 t + 5 2 t 2 + 7 t + 5
5 y 2 + 16 y + 11 5 y 2 + 16 y + 11
11 x 2 + 34 x + 3 11 x 2 + 34 x + 3
7 b 2 + 50 b + 7 7 b 2 + 50 b + 7
4 w 2 − 5 w + 1 4 w 2 − 5 w + 1
5 x 2 − 17 x + 6 5 x 2 − 17 x + 6
4 q 2 − 7 q − 2 4 q 2 − 7 q − 2
10 y 2 − 53 y − 11 10 y 2 − 53 y − 11
6 p 2 − 19 p q + 10 q 2 6 p 2 − 19 p q + 10 q 2
21 m 2 − 29 m n + 10 n 2 21 m 2 − 29 m n + 10 n 2
4 a 2 + 17 a b − 15 b 2 4 a 2 + 17 a b − 15 b 2
6 u 2 + 5 u v − 14 v 2 6 u 2 + 5 u v − 14 v 2
−16 x 2 − 32 x − 16 −16 x 2 − 32 x − 16
−81 a 2 + 153 a + 18 −81 a 2 + 153 a + 18
−30 q 3 − 140 q 2 − 80 q −30 q 3 − 140 q 2 − 80 q
−5 y 3 − 30 y 2 + 35 y −5 y 3 − 30 y 2 + 35 y
Factor Trinomials of the Form ax2+bx+cax2+bx+c using the ‘ac’ Method
In the following exercises, factor using the ‘ac’ method.
5 n 2 + 21 n + 4 5 n 2 + 21 n + 4
8 w 2 + 25 w + 3 8 w 2 + 25 w + 3
4 k 2 − 16 k + 15 4 k 2 − 16 k + 15
5 s 2 − 9 s + 4 5 s 2 − 9 s + 4
6 y 2 + y − 15 6 y 2 + y − 15
6 p 2 + p − 22 6 p 2 + p − 22
2 n 2 − 27 n − 45 2 n 2 − 27 n − 45
12 z 2 − 41 z − 11 12 z 2 − 41 z − 11
60 y 2 + 290 y − 50 60 y 2 + 290 y − 50
6 u 2 − 46 u − 16 6 u 2 − 46 u − 16
48 z 3 − 102 z 2 − 45 z 48 z 3 − 102 z 2 − 45 z
90 n 3 + 42 n 2 − 216 n 90 n 3 + 42 n 2 − 216 n
16 s 2 + 40 s + 24 16 s 2 + 40 s + 24
24 p 2 + 160 p + 96 24 p 2 + 160 p + 96
48 y 2 + 12 y − 36 48 y 2 + 12 y − 36
30 x 2 + 105 x − 60 30 x 2 + 105 x − 60
Factor Using Substitution
In the following exercises, factor using substitution.
x 4 − 6 x 2 − 7 x 4 − 6 x 2 − 7
x 4 + 2 x 2 − 8 x 4 + 2 x 2 − 8
x 4 − 3 x 2 − 28 x 4 − 3 x 2 − 28
x 4 − 13 x 2 − 30 x 4 − 13 x 2 − 30
( x − 3 ) 2 − 5 ( x − 3 ) − 36 ( x − 3 ) 2 − 5 ( x − 3 ) − 36
( x − 2 ) 2 − 3 ( x − 2 ) − 54 ( x − 2 ) 2 − 3 ( x − 2 ) − 54
( 3 y − 2 ) 2 − ( 3 y − 2 ) − 2 ( 3 y − 2 ) 2 − ( 3 y − 2 ) − 2
( 5 y − 1 ) 2 − 3 ( 5 y − 1 ) − 18 ( 5 y − 1 ) 2 − 3 ( 5 y − 1 ) − 18
Mixed Practice
In the following exercises, factor each expression using any method.
u 2 − 12 u + 36 u 2 − 12 u + 36
x 2 − 14 x − 32 x 2 − 14 x − 32
r 2 − 20 r s + 64 s 2 r 2 − 20 r s + 64 s 2
q 2 − 29 q r − 96 r 2 q 2 − 29 q r − 96 r 2
12 y 2 − 29 y + 14 12 y 2 − 29 y + 14
12 x 2 + 36 y − 24 z 12 x 2 + 36 y − 24 z
6 n 2 + 5 n − 4 6 n 2 + 5 n − 4
3 q 2 + 6 q + 2 3 q 2 + 6 q + 2
13 z 2 + 39 z − 26 13 z 2 + 39 z − 26
5 r 2 + 25 r + 30 5 r 2 + 25 r + 30
3 p 2 + 21 p 3 p 2 + 21 p
7 x 2 − 21 x 7 x 2 − 21 x
6 r 2 + 30 r + 36 6 r 2 + 30 r + 36
18 m 2 + 15 m + 3 18 m 2 + 15 m + 3
24 n 2 + 20 n + 4 24 n 2 + 20 n + 4
4 a 2 + 5 a + 2 4 a 2 + 5 a + 2
x 4 − 4 x 2 − 12 x 4 − 4 x 2 − 12
x 4 − 7 x 2 − 8 x 4 − 7 x 2 − 8
( x + 3 ) 2 − 9 ( x + 3 ) − 36 ( x + 3 ) 2 − 9 ( x + 3 ) − 36
( x + 2 ) 2 − 25 ( x + 2 ) − 54 ( x + 2 ) 2 − 25 ( x + 2 ) − 54
Writing Exercises
Many trinomials of the form x2+bx+cx2+bx+c factor into the product of two binomials (x+m)(x+n).(x+m)(x+n). Explain how you find the values of m and n.
Tommy factored x2−x−20x2−x−20 as (x+5)(x−4).(x+5)(x−4). Sara factored it as (x+4)(x−5).(x+4)(x−5). Ernesto factored it as (x−5)(x−4).(x−5)(x−4). Who is correct? Explain why the other two are wrong.
List, in order, all the steps you take when using the “ac” method to factor a trinomial of the form ax2+bx+c.ax2+bx+c.
How is the “ac” method similar to the “undo FOIL” method? How is it different?
Self Check
ⓐ After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.
ⓑ After reviewing this checklist, what will you do to become confident for all objectives?