1) y^2 - 10y + 24 11) 3x^2 + 19x + 6
2) m^2 + 11m + 24 12) 6x^2 - 5x -4
3) x^2 + 7x + 12 13) 2x^2 + 5x -7
4) x^2 + 7x + 10 14) 5x^2 - x - 6
5) x^2 -3x -10 15) 12x^2 + 7x + 1
6) x^2 - 6x + 8 16) 3x^2 + 17x -6
7) 6x^2 + 7x + 2 17) 8x^2 + 6x + 1
8) 8x^2 - 6x + 1 18) 2x^2 + x - 15
9) 2x^2 + 11x + 5 19) 4x^2 + 23x + 15
10) 7x^2 - 6x - 1 20) 2x^2 -7x -4
Recall the approach ax^2 + bx + c ( ) ( )
3x^2 -10x - 8
Step 1 Multiply a by c ac 3 x -8 = -24
Step 2 Write all combinations to get ac __ x __ -1 x 24 1 x -24
-2 x 12 2 x -12
-3 x 8 3 x -8
-4 x 6 4 x -6
Step 3 For each in step2, add the number to get b (-10) 23 -23
10 -10 got it
Step 4 Rewrite question using the two terms for middle term
3x^2 + 2x - 12x - 8
Step 5 Factorize
3x^2 + 2x -12x - 8
x(3x + 2) - 4 (3x + 2)
(3x + 2) (x - 4)
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So here's my approach:
ReplyDeletey^2 - 10y + 24:
this is in the form ax^2 + bx + c
where a = 1, b = -10 and c = 24.
firstly i list the following:
a*c = 1*24 = 24 and,
b = -10
i then say:
two number multiplied by eachother wud give 24 and the same two numbers added to eachother wud give -10.
i found these two numbers to be -6 and -4
since:
-6 * -4 = 24 and,
(-6) + (-4) = -10
using these two number i then break up the initial equation to form
y^2 - 6y - 4y + 24
i then break these up by putting terms into brackets to get
(y^2 - 6y)(- 4y + 24)
i can them factorize this. by factorization i get:
y(y-6) -4(y-6)
my next step if my final factorization to give:
(y-4(y-6)
solution to number 7:
ReplyDelete6x^2 + 7x + 2
a = 6, b = 7 and c = 2
a*c = 6*2 = 12
b = 7
since 3*4 = 12
and 3+4 = 7
then:
6x^2 + 3x + 4x + 2
(6x^2 + 3x) (+ 4x + 2)
factorize:
3x(2x + 1) +2(2x + 1)
final factorization:
(3x + 2)(2x + 1)
solution to number 13:
ReplyDelete2x^2 + 5x -7
a = 2, b = 5 and c = -7
a*c = -14
b = 5
since -7 * 2 = -14
and -7 + 2 = 5
then:
2x^2 - 7x + 2x - 7
(2x^2 - 7x)(+ 2x - 7)
factorize:
x(2x-7) +1(2x-7)
final factorization:
(x+1)(2x-7)
(y^2-10y+24 11) 3x^2 + 19x + 6
ReplyDeletey^2 - 10y + 24
The I take factors of 24
8 & 3
-1 & 24
-12 & 2
Beccause - 12 and 2 give me -10 I would factorise using these terms.
(y -12)(y + 2)
I suddenly realised that I made a mistake becaus ewhen I multipy -12 by 2 I get a negative 24. That is not correct at all.
ReplyDeleteSO Factors of 24
-1 & 24
-24 & 1
-12 & 2
-2 & 12
-6 & 4
-4 & 6
I would use -6 and -4
From there I put it into the expression:
y^2 - 6y - 4y + 24
groupiing like terms now I say:
( y -6) (y - 4)
3x^2 + 19 x + 6
ReplyDeleteThe expression is recognised as a quadratic equation because I saw that the first term was squared.
using the quadratic formula -b+- the root of b^2 - 4ac / 2a
Where a = 3
b =19
c=6
SUbstituting into the formula I get -19=- root 19^2 - 4 *3*6/2*3= - 19 root of 361 - 72 / 6 = x =19 + 289/6 = 308/6 = 51 1/3 or -19 -289/6 =-308/6 = -51.3
m^2+ 11m + 24
ReplyDeletefactors of 24
8 &3
I would use these factors because when added they give me a positive 11
factorising out I get
( m + 8) ( m + 3)
m(m+8) ( m + 3)
I am not quite sure whether number is one question to work out together or if it is 2 seperate parts. Can someone assist.
ReplyDeleteThis comment has been removed by the author.
ReplyDeleteWhen I say number I mean for example number 1
ReplyDelete6x^2 - 5x -4
ReplyDeletemultiply a by c
where a = 6 and c = -4
I would get -24
writing all the combinations to get ac
6 and -4
1 and -24
12 and -2
-8 and 3
I would use -8 and 3 because when multipied they give -24 which is c and when added they give me -5 which is b
Putting it into the expression I get
6x^2 -8x + 3 - 4
After which I would factorise
6x^2 -8x + 3 - 4
( 3x - 8) ( 2x + 3)
Taking from 6x^2 - 8x + 3 -4
ReplyDeleteFactorising out
3x(2x -8) (2x + 3)
solution to number 3
ReplyDeletex^2+7x+12
Form of Ax^2+Bx+C,
A=1, B= 7, C=12
Step1 = Find AC = 1x12 = 12
Step2 = All Cominations of 12
1x12, 2x6, 3x4
Step3 = For each in step2 add the number to get B=7.
Sol:3+4
Step4 = Rewrite the question using the two terms.
x^2+3x+4x+12
Step5 = Factorize
x(x+3)+4(x+3)
(x+3)(x+4)
solution to number 16
ReplyDelete3x^2+17x-6
Form of Ax^2+Bx+C,
A=3, B= 17, C=-6
Step1 = Find AC = 3x-6 = -18
Step2 = All Cominations of -18
1x-18, 2x-9, 3x-6
Step3 = For each in step2 add the number to get B=17x.
Sol:18-1=17
Step4 = Rewrite the question using the two terms.
3x^2+18x-x-6
Step5 = Factorize
3x(x+6)-(x-6)
(3x-1)(x+6)
x^2 + 7x + 12 13)
ReplyDeleteSomeone help me out with this please I am stuck with 12 13 how to proceed
y^2-10y+24
ReplyDelete(y-12) (y+2)
When -12+2=10 it is wise to
use -12 and 2 as the factors
m^2+11m+24
ReplyDelete(m+8) (m+3)
I chose 8 and 3 because when i add them
i get 11 and when i multiply them
i get 24
x^2+7x+12
ReplyDelete(x+4) (x+3)
x^2+7x+10
ReplyDelete(x+5) (x+2)
x^2-3x-10
ReplyDelete(x-5) (x+2)
This comment has been removed by the author.
ReplyDeletey^2-10y+24
ReplyDelete24=-1*-24
=-2*-12
=-3*-8
=-4*-6
-4+-6=-10
therefore
(y-4)(y-6)=y^2-10y+24
(3) x^2+7x+12
ReplyDelete12=1*12
=2*6
=3*4
3+4=7
therefore
x^2+7x+12=(x+3)(x+4)
(5)x^2-3x-10
ReplyDelete-10=1*-10
=2*-5
2+-5=-3
therefore
x^2-3x-10=(x-2)(x+5)
(8)8x^2-6x+1
ReplyDelete8=-1*-8
=-2*-4
-2+-4=-6
therefore
8x^2-6x+1=(4x-1)(2x-1)
(11)3x^2+19x+6
ReplyDelete18=1*18
1+18=19
therefore
3x^2+19x+6=(3x+1)(x+6)
(14)5x^2-x-6
ReplyDelete-30=1*-30
=2*-15
=3*-6
=5*-6
5+-6=-1
therefore
5x^2-x-6=(x-1)(5x-6)
(18)2x^2+x-15
ReplyDelete-30=-1*30
=-2*15
=-3*10
=-5*6
-5+6=1
therefore
2x^2+x-15=(2x+5)(x-3)