7(2y² − y + 10y − 5) = 14y² + 63y − 35 // Add like terms and multiply.
14y² + 63y − 35 = 14y² + 63y − 35 // Evaluate.
True
Q.E.D.
*Note: The AC trick is used give us a hint of how the quadratic trinomial will be factored, by giving us a new quadratic trinomial which is unequal to the original one but easier to factor. Where the original trinomial was ax² + bx + c, the new one is x² + bx + ac. We return to the original equation by “pulling the a-term back ” through a bit of trial-and-error. Usually the a-returns by dividing one of the constants which is a multiple of it and multipling the opposite coefficient by it. Though sometimes this has to be done twice or more with constants that just share factors.
the respond is b, (x – 5)(x – 7). you could examine this utilising the FOIL (First, inner, Outer, final) approach. Multiply the 1st words (x and x), the internal words (-5 and x), the outer words (x and -7), and the final words (-5 and -7). This produces x^2 -5x -7x + 35, which, after combining like words, is x^2 – 12x + 35. as a result, the expressions are the comparable. =)
Answers & Comments
Verified answer
Trial and error:
14 = 2 x7 or 14 x1
35 = 35 x 1 or 7x5
The last number is negative so in the equation there has to be one + and one negative.
(7y -5) (2y + 7)
(14y^ +49y -10y -35) doesn't work
(14y + 5) (y -7)
(14y^2 - 98y + y -35) doesn't work
All three numbers are divisible by 7.
7(2y^2 + 9y -5)
7(2y -1) (y +5)
check:
7 (2y^2 +10y -1y -5)
7(2y^ +9y - 5)
Yep that works
14y² + 63y − 35 = // Factor out 7.
7(2y² + 9y − 5) // Use the AC-trick.
AC-trick*.
y² + 9y − 10 = // Factor
(y + 10)(y − 1)
Original problem.
7(2y² + 9y − 5) =
7(y + 5)(2y − 1)
Check:
7(y + 5)(2y − 1) = 14y² + 63y − 35 // FOIL.
7(2y² − y + 10y − 5) = 14y² + 63y − 35 // Add like terms and multiply.
14y² + 63y − 35 = 14y² + 63y − 35 // Evaluate.
True
Q.E.D.
*Note: The AC trick is used give us a hint of how the quadratic trinomial will be factored, by giving us a new quadratic trinomial which is unequal to the original one but easier to factor. Where the original trinomial was ax² + bx + c, the new one is x² + bx + ac. We return to the original equation by “pulling the a-term back ” through a bit of trial-and-error. Usually the a-returns by dividing one of the constants which is a multiple of it and multipling the opposite coefficient by it. Though sometimes this has to be done twice or more with constants that just share factors.
14y^2 + 63y - 35
= 7(2y² + 9y – 5)
= 7(2y² + 10y – y – 5)
= 7[2y(y + 5) –1( y + 5)]
= 7(2y – 1)( y + 5)
----------------
the respond is b, (x – 5)(x – 7). you could examine this utilising the FOIL (First, inner, Outer, final) approach. Multiply the 1st words (x and x), the internal words (-5 and x), the outer words (x and -7), and the final words (-5 and -7). This produces x^2 -5x -7x + 35, which, after combining like words, is x^2 – 12x + 35. as a result, the expressions are the comparable. =)
7(2y-1)(y+5)
7(2y-1)(y+5)
7(2y^2+9y-5)
7(2y-1)(y+5)